MHP Resource Set 2 - teachers choice software Set 2.pdf · Introduction The Maths Helper Plus Resource Set 2 consists of 25 classroom mathematics activities in trigonometry, along

Maths Helper Plus Resource Set 2 25 interactive trigonometry activities for

the Maths Helper Plus software.

Teachers’ Choice Software

Maths Helper Plus Resource Set 2 - Trigonometry Copyright © Bruce A. Vaughan 2003

In the following paragraphs, ‘Material’ refers to ‘Maths Helper Plus Resource Set 2’, and includes the activity sheets and Maths Helper Plus document files in the resource set, being in electronic or printed form. Permission is granted for the purchaser to copy the material, but only for non-commercial educational purposes of the purchasing individual or institution. Under no circumstances is this permission transferable to another party. All rights are reserved. Except under the conditions outlined above, and those specified in the Copyright Act 1968 of Australia and subsequent amendments, no part of the Material may be reproduced, stored in a retrieval system or be broadcast or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise.

Bruce A. Vaughan Teachers’ Choice Software, P.O. Box 247, Goodna Qld 4300 Australia Telephone: (07) 3288 5035 Facsimile: (07) 3288 3702 E-mail: [email protected] Internet: www.teacherschoice.com.au Printed 2003

Introduction

The Maths Helper Plus Resource Set 2 consists of 25 classroom mathematics activities in trigonometry, along with supporting Maths Helper Plus files. This resource set makes use of many of the powerful new features in version 2 of Maths Helper Plus. I stress that these are activities, not lessons. I leave it to the teacher to decide how, when and where they are used. They are suitable as a basis for lessons, demonstrations, group work, projects and student assignments. In general, these activities are self correcting, so no solutions are required. The computer provides immediate feedback so that errors are corrected when they occur. Some open ended questions may require teacher input. The Maths Helper Plus support files are needed by the various activities. These must be made available on the network or hard drive of the users’ computer wherever the Resource Set is going to be used. It is recommended that these files be made ‘read only’ so that students cannot spoil the originals. Before attempting the activities with your class, make sure you know how to locate and open the Maths Helper Plus support files for this resource set. They must not be confused with the support files from other resource sets. To help prevent such confusion, the name of each support file for this resource set begins with: ‘R2 -’. Normally, these files will be located in a folder called: ‘MHP Resource Set 2’. The worksheets assume little prior knowledge of Maths Helper Plus, but I do recommend that students work through the ‘Basic Skills’ activity sheet before attempting the other worksheets. Lastly, this resource set requires Maths Helper Plus version 2 or greater to be installed on your computer system. Enjoy!

Contents Note: The ‘time’ and ‘year’ columns below are meant to indicate the time for students to complete an activity, as well as the minimum year level appropriate for an activity. These are only wild guesses, reality may prove very different! Title Description Resource files Time Yr Pge Basic Skills - Activity 1 - Basic skills for using Maths Helper Plus nil 30 min 8+ 1 Trigonometry - Activity 1

Angle Relationships - Sum of interior angles = 180º - Exterior angle = sum of interior opposite angles

R2 - Angle relationships 1.mhp

35 min 8+ 3

Activity 2 - Sides opposite congruent angles are congruent - Angles opposite congruent sides are congruent - Sides and angles in equilateral triangles

R2 - Angle relationships 2.mhp

30 min 8+ 5

Activity 3 Congruency - Naming corresponding angles and sides in

congruent triangles. - Statements of congruency

R2 - Congruency 1.mhp

25 min 9+ 7

Activity 4 - Demonstrating the failure of AAA and ASS to guarantee congruency.

- Establishing that RHS guarantees congruency.


40 min 9+ 9

Activity 5 - Interpreting triangle diagrams to establish congruency relationships.


40 min 9+ 11

Activity 6 - Interpreting triangle diagrams to establish congruency relationships.

- Applying congruency rules.

R2 - Congruency 5.mhp R2 - Congruency 6.mhp

40 min 9+ 13

Activity 7 Pythagoras - Demonstrating Pythagoras’ Theorem - Pythagorean triples

R2 - Pythagoras 1.mhp

40 min 9+ 15

Activity 8 - Using Pythagoras’ Theorem to solve for an unknown hypotenuse or leg of a right triangle.

R2 - Pythagoras 2.mhp

40 min 10+ 17

Activity 9 Tangent Ratio (right triangles) - Defining the terms: ‘opposite’ and ‘adjacent’. - Defining the tangent ratio. - Demonstrating that the tangent ratio depends only

on the angle, not the size of the triangle

R2 - Tangent ratio 1.mhp

35 min 9+ 19

Activity 10 - Making a table of tangent ratios for common angles.

- Using known tangent ratios to solve for unknown angles in right triangles.


40 min 9+ 21

Activity 11 - Using known tangent ratios to solve for unknown opposite or adjacent sides in right triangles.


40 min 10+ 23

Activity 12 - Explanation of angles of elevation and depression.

- Solving word problems involving angles of elevation and depression.


40 min 10+ 25

Activity 13 Sine Ratio (right triangles) - Defining the sine ratio. - Identifying the opposite side and the hypotenuse

in right angled triangles. - Calculating sine ratios from triangle diagrams. - Establishing that the sine ratio does not depend on

the size of a triangle.

R2 - Sine ratio 1.mhp 40 min 9+ 27

Activity 14 - Making a table of sine ratios for common angles. - Using known sine ratios to solve for unknown

opposite angles in right triangles. - Using known sine ratios to solve for unknown

sides in right triangles. ( Both opposite and hypotenuse. )


Activity 15 - Using known sine ratios to solve for unknown legs in right triangles.

- Using known sine ratios to solve for an unknown hypotenuse in a right triangle.


Title Description Resource files Time Yr Pge Activity 16 Cosine Ratio (right triangles)

- Defining the cosine ratio. - Identifying the adjacent side and the hypotenuse

in right angled triangles. - Calculating cosine ratios from triangle diagrams. - Establishing that the cosine ratio does not depend

on the size of a triangle.

R2 - Cosine ratio 1.mhp

40 min 9+ 33

Activity 17 - Making a table of cosine ratios for common angles.

- Using known cosine ratios to solve for unknown adjacent angles in right triangles.


40 min 9+ 35

Activity 18 - Using known cosine ratios to solve for unknown legs in right triangles.

- Using known cosine ratios to solve for an unknown hypotenuse in a right triangle.


40 min 10+ 37

Activity 19 Similar Triangles - Defining ‘similar’ triangles. - Locating corresponding angles and sides. - Calculating scale factors. - Experimenting with scale factors and triangle

shape for pairs of similar triangles.

R2 - Similarity 1.mhp 40 min 9+ 39

Activity 20 - Finding unknown corresponding angles in similar triangles.

- Finding unknown corresponding sides in similar triangles.

R2 - Similarity 1.mhp 40 min 10+ 41

Activity 21 General Triangle Solution ( Type 1, Given three sides. ) - Procedure outlined in detail. - Six practice problems. Use Maths Helper Plus to

generate solution and to correct the working.

R2 - Triangle_solver_SSS.mhp

50 min 11+ 43

Activity 22 ( Type 2, Given two sides and included angle. ) - Procedure outlined in detail. - Six practice problems. Use Maths Helper Plus to


R2 - Triangle_solver_SAS.mhp

50 min 11+ 45

Activity 23 ( Type 3, Given two sides and non-included angle.) - Demonstrate the ambiguous case. - Procedure outlined in detail. - Three practice problems. Use Maths Helper Plus

to generate solution and to correct the working.

R2 - Triangle_solver_ASS.mhp

50 min 11+ 47

Activity 24 ( Type 4, Given two angles and one side.) - Procedure outlined in detail. - Six practice problems. Use Maths Helper Plus to


R2 - Triangle_solver_AAS.mhp

50 min 11+ 49

Activity 25 ( Type 5, Given three (x,y) points. ) - Procedure outlined in detail. - Six practice problems. Construct and solve

triangles. Use Maths Helper Plus to generate solution and to correct the working.

R2 - Triangle_solver_ASS.mhp

50 min 11+ 51

INTENTIONALLY BLANK PAGE

Name: Class: p 1

Maths Helper Plus Resource Set 2. Copyright © 2003 Bruce A. Vaughan, Teachers’ Choice Software

Basic Skills - Activity 1 Basic skills for using Maths Helper Plus

‘Maths Helper Plus’ is a computer program for doing high school maths. These are some skills you need in order to use the program properly. Tick the boxes: to indicate which skills you have completed. Skill 1 - Starting Maths Helper Plus: Your teacher will show you how to start Maths Helper Plus on your computer system. A common way to start it is by double clicking on the program icon:

• Practice starting Maths Helper Plus now. Skill 2 - Moving the ‘splitter bar’: When Maths Helper Plus starts, it displays two windows called the ‘text view’ and the ‘graph view’. The two windows are separated by the ‘splitter bar’. (See below)

Text View Graph View

Splitter Bar

Sometimes you only want to see the text view or the graph view. Other times you want to see some of each. Method 1 There are keyboard shortcuts for moving the splitter bar:

‘Ctrl+T’ means to hold down a ‘Ctrl’ key while you press the ‘T’ key. This displays the text view only. Ctrl+G displays the graph view only, and Ctrl+B displays both views.

Method 2 Move the mouse cursor over the splitter bar. When the shape of the cursor changes to a horizontal arrow, click and drag with the mouse. Release the mouse button when the splitter bar is where you want it.

• Try the keyboard and the mouse methods for moving the splitter bar. Skill 3 - Entering your data: Maths Helper Plus can accept many kinds of mathematical information, called ‘input data’. Three common types of input data are: (x,y) points, statistical data, and equations, You enter your data into the input box on the text view: I N P U T B O X To activate the input box for editing, either press the F3 key, OR click inside the input box with the mouse.

NOTE: If the input box is already selected (see skill 4 on the next page) no F3 or mouse clicking is needed. Just start typing! Follow these steps to type these points: (1,2) (2,1) (4,4) into Maths Helper Plus:

1. Press the F3 key The ‘short cuts’ box will appear. You can click on the shortcut buttons to type special symbols. If it displays itself in an inconvenient place, point to the words: ‘Shortcut box’ (at its top) with the mouse, then click and drag it somewhere out of the way.

2. Type your data 3, Press the Enter key

The graph view displays dots for the points on the graph, and the text view displays the data you typed in a double edit box:

(1,2) (2,1) (4,4) Your input data is displayed here.

You can type a descriptive label here.

To type further data sets, you don’t need to press the F3 key each time. As long as you used the ‘Enter’ key to complete your last data entry, the input box will be selected automatically, ready for your next entry. You can press ‘Enter’ again to display the shortcuts box, or just start typing your next data if you don’t care about the shortcuts box! Create four more data sets using the input data given below:

Type of data set: Input Data Graph view display: (x,y) points (1,0) (2,1) (3,2) (4,3) (5,4) Plotted points. equation y = x − 1 Graph of the equation. equation y = −x + 1 Graph of the equation. Set of statistical data 1;2;1;4;3;1;2;4;1;1;2 Histogram plot of the data.

Skill 4 - Selecting a data set Expand

box An ‘expand box’ is attached to every data set on the text view: The selected data set has a yellow expand box.

• Find the selected data set on your text view. To select a data set, you click beside its double edit box on the text view:

• Select a different data set by clicking beside it on the text view. Skill 5 - Selecting (x,y) point and equation data sets Carefully point with the mouse to the line graph of an equation or the exact middle of a plotted point. Now click the left mouse button. This will select that data set and its expand box will turn yellow.

• Try identifying your plotted points and line graphs by clicking on them. Watch for the expand box that becomes yellow.

Skill 6 - Editing your data: To change the input data you have already entered:

Method 1: Select the data set, then either (1) press Enter to begin editing the data, or (2) To add more data at the end of existing data, begin typing without pressing Enter.

NOTE: To edit the descriptive label, hold down ‘Shift’ while you press ‘Enter’. Method 2: Click on the data, make your changes, then click outside of the edit box. Try this:

• Click on the input data: (1,2) (2,1) (4,4) • Change (2,1) to (5,2) • Click outside of the edit box

Skill 7 - Displaying the options dialog box Every data set has an options dialog box where you can choose to display graphs and calculations for your input data. The options available depend on the type of data set. To display the options dialog box:

Method 1: Select the data set, then hold down ‘Ctrl’ while you press ‘Enter’, or Method 2: Double click the mouse on the text view beside the data.

Try this: • Double click on the text view beside the statistical data set to display its options dialog box. • Select the ‘Calculations’ tab at the top of the dialog box, then click to select ‘range’. • Click OK to close the dialog box.

The text view now displays the calculations for finding the range of this data.

• Double click exactly in the centre of a plotted (x,y) point to display the options dialog for the set of points. (You can also display it by double clicking beside the data set on the text view.)

• In the ‘(x,y) Point Settings’ tab, choose some of the available options and note the effect on the graph. Click OK to close the dialog box.

Skill 8 - Deleting unwanted data sets To delete one data set you don’t want, select it as described above, then hold down a ‘Ctrl’ key while you press the ‘Delete’ key.

• Select and delete one data set.

To delete ALL data sets, you choose the ‘Delete All’ command in the ‘Edit’ menu. • Delete all remaining data sets.

(1,2) (5,2) (4,4)

(1,2) (5,2) (4,4)

Click here to select

(1,2) (5,2) (4,4)

NOTE: Usually, you can also move the selection by clicking the up arrow or down arrow keys on your keyboard, although this facility is sometimes disabled.

Name: Class: p 3


Trigonometry - Activity 1 Angle relationships: Sum of interior angles; Interior opposite angles

(a) Interior Angles of Triangles Every triangle has three interior angles:

Interior angles

You will measure and record the interior angles and their sum, for several different triangles. 1) Start Maths Helper Plus and load the ‘R2 - Angle relationships 1.mhp’ document. The graph view will display a triangle. 2) Press the F5 key to display the parameters box. (See below.) The values of ‘A’, ‘B’ and ‘C’ are the lengths of the three sides of the triangle:

A = 6B = 6

C = 660°

60°

60°

3) Change the length of the sides of the triangle, like this: Step 1: Click on the parameters box, on the right side of the value to be changed. Step 2: Press the ‘Backspace’ key to delete the old value. Step 3: Type the new value. Step 4: Click the ‘Update’ button.

click here

For each row of the table below, change the triangle sides to have the given lengths. For each triangle you make, record the three interior angles in the table in any order. Use a calculator to add the three angles and record the sum in the last column.

Side ‘A’ Side ‘B’ Side ‘C’ Angle 1 Angle 2 Angle 3 Sum of the three angles

6 6 6 60º 60º 60º 180º

3 6 6

8 6 6

10 6 6

10 8 6

10 15 6

6 8 5

6 8 11

6 8 13

6 8 14 4) Complete this sentence: ‘The sum of the interior angles of a triangle = ___________’

(b) Exterior angle relationship If any side of a triangle is extended beyond the triangle, an exterior angle is created:

‘Exterior’ angle ‘Interior opposite’ angles

How many exterior angles does a triangle have ? __________ For every exteror angle there are two interior opposite angles. (See diagram above.) By measuring an exterior angle and its interior opposite angles for several different triangles, you will be able to demonstrate the relationship between these angles. 5) Change the ‘X’ value on the parameters box from ‘1’ to ‘2’. This causes only the exterior angle and interior opposite angles to display. 6) For each row of the table below, change the triangle sides to have the given lengths.

For each triangle you make: • record the exterior angle and the two interior opposite angles in the table. • use a calculator to add the two interior opposite angles and record the sum in the last column.

interior opposite angles

Side ‘A’ Side ‘B’ Side ‘C’ exterior angle angle 1 angle 2 angle 1 + angle 2

6 6 6 120º 60º 60º 120º

3 6 6

8 6 6

10 6 6

10 8 6

10 15 6

6 8 5

6 8 11

6 8 13

6 8 14 7. Write a sentence that describes the relationship between an exterior angle of a triangle and the sum of its interior opposite angles:

Name: Class: p 5


Trigonometry - Activity 2 Angle relationships: Isosceles and equilateral triangles.

(a) Angle Relationships in Isosceles Triangles An isosceles triangle has two sides with the same length.

This diagram shows an isosceles triangle. The two equal sides are marked with short lines: / and the arrows point to the angles that are opposite to these sides. You will create several different isosceles triangles, and record the angles that are opposite to the two equal sides. 1) Start Maths Helper Plus and load the ‘R2 - Angle relationships 2.mhp’ 2) Press the F5 key to display the parameters box. (See below.) The values of ‘A’, ‘B’ and ‘C’ are the lengths of the three sides of the triangle:

A = 6B = 6

C = 660°

60°

60°

3) Change the length of the sides of the triangle, like this: Step 1: Click on the parameters box, on the right side of the value to be changed. Step 2: Press the ‘Backspace’ key to delete the old value. Step 3: Type the new value. Step 4: Click the ‘Update’ button.

click here click here

Lines or angles that are exactly the same are said to be ‘congruent’. For each row of the table below, change the triangle sides to have the given lengths. For each triangle you make, record the two angles that are opposite to the congruent sides. (The first row has been done for you. See if you agree with the values for angle 1 and angle 2.)

Side ‘A’ Side ‘B’ Side ‘C’ Angle 1 Angle 2 7 7 6 64.6231º 64.6231º

10 10 6

6 5 6

9 5 9

10 8 6

10 6 10

10 7 10 4) Complete this sentence: ‘If two sides of a triangle are congruent, the angles ____________ those sides are _____________’

(b) Angle Relationships in Equilateral Triangles An equilateral triangle has three congruent sides: You will create several different equilateral triangles, and record the three interior angles for each. 5) For each row of the table below, change the triangle sides to have the given lengths. For each triangle you create, record the three interior angles in the table below:

Side ‘A’ Side ‘B’ Side ‘C’ angle 1 angle 2 angle 3 5 5 5 60º 60º 60º

6 6 6

7 7 7

8 8 8

10 10 10 6) Complete this sentence: ‘An equilateral triangle has three c_____________ angles of _______ degrees’

Name: Class: p 7


Trigonometry - Activity 3 Congruency: Naming corresponding sides and angles; Congruency statements.

Two triangles are said to be congruent if they have exactly the same size and shape. Every triangle has three interior angles and three sides, and for two triangles to be congruent, the sides and angles of one must be the same as the sides and angles of the other. 1) Start Maths Helper Plus and load the ‘R2 - Congruency 1.mhp’ document. The graph view will display two congruent triangles similar to these: On this diagram, congruent sides and angles have matching symbols. The letters ‘A’, ‘T’, ‘P’, ‘J’, ‘E’ and ‘L’ identify the vertices of the triangles.

J

E

P

A

T

L

For congruent triangles, the vertices of one triangle are said to correspond to those in the other triangle. If the congruent triangles were overlaid on top of one another, then ‘T’ would be exactly on top of ‘E’. So ‘T’ corresponds to ‘E’. This is often written like this: T ↔ E 2) Complete the following: A ↔ _____ L ↔ _____ To be sure that the triangles are congruent, we can overlay them on the screen. 3) Press the F5 key to display the parameters box. (See below.)

• Click on the ‘A’ edit box

• Click on the slider button Use the up arrow and down arrow keys on your keyboard to move the right hand triangle so that it exactly fits on top of the left hand triangle. Are these triangles congruent ? _________ Congruency notation: This symbol: means ‘≅ is congruent to’.

So to state that our two triangles are congruent, we can say: ∆ T A P ∆ E J L ≅ When you state a congruency relationship, the letters must be in corresponding order. 4) Check this on the triangles: Does ‘T’ ↔ ‘E’ and ‘A’ ↔ ‘J’ and ‘P’ ↔ ‘L’ ? _________ 5) We can also write this congruency relationship with the letters in a different order, as long as they are still written in corresponding order. Complete these statements: (b) ∆ A T P ∆ _______ (b) ∆ ______ ∆ L E J ≅ ≅ Sides and Angles: Letters in corresponding order can also define sides and angles.

• Two letters for sides, eg ‘AT’. ( AT JE means: Side ‘AT’ is congruent to side ‘JE’. ) ≅

• Three letters for angles, eg PAT. ( So: PAT LJE means: Angle ‘PAT’ is congruent to angle ‘LJE’ ) ≅

6) Complete the statements of congruency below. In each case, study the triangles on your computer screen so that you get the vertices in corresponding order.

Sides: (a) A P ____ ≅

(b) T P ____ ≅

(c) A T ____ ≅

(d) ____ E L ≅

(e) ____ J E ≅

Angles:

(f) ATP ≅

(g) LEJ ≅

(h) ELJ ≅

(i) PAT ≅

(j) TPA ≅

Name: Class: p 9


Trigonometry - Activity 4 Minimum conditions to establish congruency.

An architect has designed a triangular outdoor shade for a school play area, to be suspended by its corners: This triangle has three sides and three interior angles. By choosing wisely, only three of these six values need to be specified in order to make sure that the shade is built correctly in the factory, but which three ? If a set of measurements can create more than one triangle, then the shade may be constructed incorrectly. To solve this problem, we will list all of the possibilities, then eliminate the combinations that will not work. A triangle has these six values: (For simplicity, we will use capital letters for the angles, and little letters for the sides.)

a

c

b

A

C

B

angles: A, B, C

sides: a, b, c

Here are the twenty possible sets of three values from this triangle. Similar combinations have been grouped:

Set of three measurements: Description Symbol

ABC Three angles AAA

ABa, ABb, ABc, ACa, ACb, ACc, BCa, BCb, BCc, Two angles and one side AAS

aBc, cAb, bCa Two sides and the angle between the sides. (The included angle) SAS

abA, Bab, cbC, Bcb, acA, Cac Two sides and an angle not between the sides.

ASS RHS

abc Three sides SSS

Investigation 1, Three angles, AAA Are the three angles of a triangle sufficient information to construct a given triangle ? 1) Start Maths Helper Plus and load the ‘R2 - Congruency 2.mhp’ document. The graph view will display this triangle: 2) Press the F5 key to display the parameters box. (See below.) 40

80

60

• Click on the ‘A’ edit box

• Click on the slider button 3) Press and hold the up arrow key

on the computer keyboard until the blue triangle stops enlarging.

4) The two triangles on your screen always have the same interior angles: 40, 60, 80. Would these three angles be enough information to build a triangular structure accurately ? _______ Why ?

5) Press and hold the down arrow key until the blue triangle is as small as possible, then use the up arrow key to return it to its original size. Put a cross: in the box near ‘AAA’ in the table on the front side of this sheet to indicate that two triangles do not have to be congruent just because their interior angles are congruent.

Investigation 2, Two sides and a non-included angle If you knew only the lengths of two sides and one non-included angle, could you build the triangular structure accurately ? ( Remember, a non-included angle does not lie between the two given sides.) 6) Load the ‘R2 - Congruency 3.mhp’ document. The graph view will display this diagram: This is a triangle under construction. We have been given three values, the two sides S1 and S2, and one angle ‘A’ that is not between the two given sides. This method has been followed so far:

(a) a horizontal baseline was drawn (see diagram)

(b) side S1 was added, making angle ‘A’ with the baseline

A

S1

S2

baseline

Now we are just about to add side S2 to join side S1 to the baseline. How many ways can we do this ?

7) If the parameters box is not already displayed, press F5 to display it. • Click on the ‘B’ edit box on the parameters box. • Click on the slider button.

8) Use the up and down arrow keys on the keyboard to rotate side S2 so that it just meets the baseline. You should be able to make two different triangles.

Complete these diagrams by drawing side S2 in the two positions you found that can make a triangle: Triangle 1 Triangle 2

A

S1

A

S1

Put a cross: in the box near ‘ASS’ in the table on the front side of this sheet to show that two triangles do not have to be congruent just because two of their sides and one non-included angle are congruent.

Investigation 3, For right triangles: Right angle, Hypotenuse and 1 other side If side S2 is only just long enough to reach the baseline, then it will only meet the baseline at one point. S2 will then make a right angle with the baseline. 9) Set up the parameters box values:

• Click on the ‘A’ edit box on the parameters box. Press ‘Backspace’ to delete the ‘8’ and type 5. • Click on the ‘B’ edit box, Backspace and change the value to 90. • Click on the ‘slider button’.

The labels will change on the diagram. The right angle will be marked, the side ‘S1’ is now the hypotenuse, and so is called: ‘H’, and ‘S2’ is now called: ‘S’. 10) Use the up and down arrow keys on your keyboard to rotate side ‘S’. You will see that only one triangle is

possible when side ‘S’ just touches the baseline. Put a tick: in the box near ‘RHS’ in the table on the front side of this sheet to show that two right triangles have to be congruent if their hypotenuses and one other side are congruent.

11) All other combinations of 3 triangle measurements are sufficient to specify a unique triangle.

Put a tick: in all of the remaining boxes in the table on the front side of this sheet.

Name: Class: p 11


Trigonometry - Activity 5 Congruency: Practice writing congruency relationships for pairs of congruent triangles.

1) Start Maths Helper Plus and load the ‘R2 - Congruency 4.mhp’ document. The triangle displayed on the graph view is really two congruent triangles superimposed. One triangle remains fixed, while you can move the other triangle to other positions. 2) Press the F5 key to display the parameters box. (See below.)

3) Learn to move the triangle. You can move one of the congruent triangles by changing the numbers in the edit boxes of the parameters box. This is how the values affect the triangle: • ‘A’ rotates the triangle. • ‘B’ moves the triangle horizontally. • ‘C’ moves the triangle vertically. • ‘D’ flips the triangle horizontally if it has a non zero value. • ‘X’ flips the triangle vertically if it has a non zero value. To rotate or move the triangle, click on the edit box for the movement you want ( ‘A’, ‘B’ or ‘C’ ), then drag the slider up and down with the mouse.

• Try rotating and moving the triangle. To flip the triangle, click on the ‘D’ or ‘X’ edit box as required, then press the Backspace key to delete the old value. Now type ‘1’ and click the ‘Update’ button. Change the value back to ‘0’ to undo.

• Try flipping the triangle horizontally and vertically, then undo. Hints for using the parameters box:

(a) Always click the edit box you want to change before dragging the slider. (b) If you are dragging the slider and want finer control, use the up and down keyboard arrow keys. (c) If the slider reaches the top or bottom but you want to go further, click on the edit box again. (d) To enter a particular value, click on the edit box, change the value and then click ‘Update’.

Now return the triangle to its original position.

• To do this, click on each edit box, change the value to 0, then click Update. Corresponding corners of the triangles have the same shaped markers drawn on them, and corresponding sides have the same colour. 4) For each pair of congruent triangles below, write the congruency relationship for the triangles, then use the moving triangle on your screen to duplicate the diagram and correct your answer. (a) Complete the statement of congruency for these congruent triangles:

∆ U T V ∆ _______ ≅

edit boxes

update button

slider

A C

V

B

U

T

C A

Use Maths Helper Plus to correct your answer: • Set up the moving triangle with these values: A = 0, B = 2.88, C = 0.8, D = 1, X = 0 • Follow the dotted lines between the triangles to find corresponding vertices. • Compare with your answer and correct any mistakes.

(b) Complete the statement of congruency for these congruent triangles: ∆ Z W Y ∆ _______ ≅

Use Maths Helper Plus to correct your answer: • Set up the moving triangle with these values: A = 1.28, B = -1.92, C = 1.4, D = 1, X = 0 • Follow the dotted lines between the triangles to find corresponding vertices. • Compare with your answer and correct any mistakes.

(c) Complete the statement of congruency for these congruent triangles:

X

P

Q

A

B

Z

Y

W

∆ P Q B ∆ _______ ≅

Use Maths Helper Plus to correct your answer: • Set up the moving triangle with these values: A = -3.14, B = 3, C = 0.32, D = 0, X = 0 • Follow the dotted lines between the triangles to find corresponding vertices. • Compare with your answer and correct any mistakes.

(d) Complete the statement of congruency for these congruent triangles:

M

K L

∆ M L J ∆ _______ ≅

J Use Maths Helper Plus to correct your answer: • Set up the moving triangle with these values: A = 0, B = 0.96, C = -0.36, D = 1, X = 0 • Follow the dotted lines between the triangles to find corresponding vertices. • Compare with your answer and correct any mistakes.

(e) Complete the statement of congruency for these congruent triangles: H

G

∆ P J G ∆ _______ ≅

P

J Use Maths Helper Plus to correct your answer:

• Set up the moving triangle with these values: A = 0, B = 0, C = -2.52, D = 0, X = 1 • Follow the dotted lines between the triangles to find corresponding vertices. • Compare with your answer and correct any mistakes.

Name: Class: p 13


Trigonometry - Activity 6 Establishing congruency, and more practice writing congruency relationships for pairs of congruent triangles.

IMPORTANT: This activity assumes that you have completed at least the front side of ‘Trigonometry Activity 5’. Right angled triangles... 1) Start Maths Helper Plus and load the ‘R2 - Congruency 5.mhp’ document. The right triangle displayed on the graph view is really two congruent triangles superimposed. One triangle remains fixed, while you can move the other triangle to other positions.

See ‘Trigonometry Activity 5’ to find out how to change the triangle position. 2) If not displayed already, press the F5 key to display the parameters box. NOTE: Each of the following questions contains a pair of congruent triangles, however, when stating the reason for congruency in part (a) of the questions, base your reason only on details marked on the diagrams. 3) (a) Tick one of the following options as the reason for congruency of these triangles:

T

10 26

P S

A

10 26

J

AAS SAS RHS SSS

(b) Complete this statement of congruency

W ∆ A P S ∆ _______ ≅

(c) Use Maths Helper Plus to correct your answer: • Set up the moving triangle with these values: A = 0, B = 3.36, C = 2.4, D = 1, X = 0 • Follow the dotted lines between the triangles to find corresponding vertices. • Compare with your answer and correct any mistakes.

4) (a) Tick one of the following options as the reason for congruency of these triangles:

L AAS SAS RHS SSS

8 8 M

N

10 (b) Complete this statement of congruency

∆ L N B ∆ _______ ≅B

(c) Use Maths Helper Plus to correct your answer:

• Set up the moving triangle with these values: A = 0, B = -4.28, C = -0.64, D = 1, X = 0 • Follow the dotted lines between the triangles to find corresponding vertices. • Compare with your answer and correct any mistakes.

5) (a) Find two congruent triangles in this diagram, then tick one of the following options as the reason for congruency:

AAS SAS RHS SSS (b) Complete this statement of congruency:

∆ _______ ∆ _______ ≅ (c) Use Maths Helper Plus to correct your answer:

• Set up the moving triangle with these values: A = 1.57, B = 0.36, C = 0.64, D = 0, X = 1

• Follow the dotted lines between the triangles to find corresponding vertices.

C

J

T

P

A

W

• Compare with your answer and correct any mistakes. Obtuse angled triangles... Load the ‘R2 - Congruency 6.mhp’ document for obtuse angled triangles. 6) (a) Find two congruent triangles in this diagram, then tick one of the following options as the reason for congruency:

AAS SAS RHS SSS (b) Complete this statement of congruency:

∆ _______ ∆ _______ ≅ (c) Use Maths Helper Plus to correct your answer:

• Set up the moving triangle with these values: A = 0, B = 0, C = -3, D = 1, X = 1

• Follow the dotted lines between the triangles to find corresponding vertices.

O

N

L

K M

• Compare with your answer and correct any mistakes. 7) (a) Find two congruent triangles in this diagram with vertices at the letter names, then tick one of the following

options as the reason for congruency. (Similar shapes indicate congruent lines.)

R

T N

I AAS SAS RHS SSS (b) Complete this statement of congruency

∆ _______ ∆ _______ ≅

(c) Use Maths Helper Plus to correct your answer: • Set up the moving triangle with these values:

A = 0, B = -0.64, C = 0.64, D = 1, X = 0 A

• Follow the dotted lines between the triangles to find corresponding vertices. • Compare with your answer and correct any mistakes.

p 15 Name: Class:


Trigonometry - Activity 7 Pythagoras: Visual demonstration of the rule; Pythagorean triples.

The rule of Pythagoras applies to right angled triangles. It can be used to find an unknown side of a right angled triangle, or to prove that a given triangle is right angled.

legs

hypotenuse A right angled triangle, or ‘right’ triangle, contains a right angle: The longest side in a right angled triangle is called the ‘hypotenuse’. The hypotenuse is always opposite to the right angle. The two other sides are shorter than the hypotenuse, and are called ‘legs’. The rule of Pythagoras states that in any right angled triangle with legs ‘a’ and ‘b’ and hypotenuse ‘c’,

b

a

c

c2 = a2 + b2 NOTE: It follows that for any triangle with side lengths ‘a’, ‘b’ and ‘c’, if c2 = a2 + b2, then that triangle is a right angled triangle. Part a: Demonstrating the rule of Pythagoras using areas. To demonstrate the rule of Pythagoras we will use the fact that ‘a2’ is the area of a square with side ‘a’, and similarly for ‘b’ and ‘c’. This diagram shows the sides and areas: It follows from the rule of Pythagoras that: area3 = area1 + area2 You will now demonstrate the rule of Pythagoras using areas. 1) Start Maths Helper Plus and load the ‘R2 - Pythagoras 1.mhp’ document. A right angled triangle diagram will be displayed. 2) Press the F5 key to display the parameters box. (See below.) The values of ‘A’ and ‘B’ are the lengths of the two legs ‘a’ and ‘b’ of the right triangle:

area = a2 1

area = c2 3

area = b2 2

c

b

a

Side ‘a’

slider Side ‘b’

3) Change the length of the legs of the triangle, like this: Step 1: Click on the parameters box, on the right side of the value to be changed. Step 2: Press the ‘Backspace’ key to delete the old value. Step 3: Type the new value. Step 4: Click the ‘Update’ button.

click here

4) For each row of the table below: (a) calculate a2, b2 and a2 + b2 without using the computer. Write these value in the table. (b) enter the ‘a’ and ‘b’ values into the parameters box as shown above, and click ‘Update’.

NOTE: If the diagram becomes too big for your computer screen, press the ‘F10’ key to reduce its size. To make the diagram bigger, hold down ‘Shift’ while you press ‘F10’. (c) correct your values of a2 and b2. Copy the ‘c’ and c2 values from the computer screen to the table.

a a2 b b2 a2 + b2 c2 c 3 4

2 4

5 3

5 6

6 1 If the rule of Pythagoras is true, then the a2 + b2 and c2 columns of the table should be equal. Check this. Part b: Pythagorean Triples A set of three integers: { a, b, c } is a ‘Pythagorean triple’ if a2 + b2 = c2. Three integers will only form the lengths of the sides of a right angled triangle if they are a Pythagorean triple. For example, { 3, 4, 5 } is a Pythagorean triple because 32 + 42 = 52 , so we can make a right angled triangle with these side lengths. To see if three numbers are a Pythagorean triple:

(a) The two smallest numbers are ‘a’ and ‘b’. Find the sum: a2 + b2 of the squares of the smallest two of the three numbers.

(b) The largest number is ‘c’. Calculate c2.

(c) If a2 + b2 = c2 then the integers: { a, b, c } form a Pythagorean triple. 5) For each set of three integers in the table below,

(a) Write the ‘a’, ‘b’ and ‘c’ values in the table. ( As long as ‘a’ and ‘b’ are the two smallest numbers, you can decide which number is ‘a’ and which is ‘b’. )

(b) Calculate the values for the other columns of the table and write your answers in the spaces provided.

(c) If the a2 + b2 and c2 values are the same, then these numbers are a Pythagorean triple. Write ‘yes’ in the last column of the table if this is true, or ‘no’ if it is false.

(d) Enter your ‘a’ and ‘b’ values into Maths Helper Plus as in question ‘4’ above.

NOTE: If the diagram becomes too big for your computer screen, press the ‘F10’ key to make it smaller. To make the diagram bigger, hold down ‘Shift’ while you press ‘F10’.

Check your values of a2, b2 and c2 with those calculated by Maths Helper Plus.

Set of integers a a2 b b2 c c2 a2 + b2 Pythagorean

Triple ( yes/no) 3, 5, 4

2, 3, 4

13,5,12

24,25,7

11, 9, 4

8, 15, 17

Name: Class: p 17


Trigonometry - Activity 8 Pythagoras: Solving for unknown sides of right triangles.

In this activity, you will review the steps for finding unknown sides of right angled triangles using the rule of Pythagoras. You will then use these techniques to solve several triangle problems and use Maths Helper Plus to check you working steps. The rule of Pythagoras states that in any right angled triangle with legs ‘a’ and ‘b’ and hypotenuse ‘c’,

b

a

c

c2 = a2 + b2 This rule can be used (a) to find the hypotenuse, given the two leg lengths, or (b) to find an unknown leg length, given the other leg length and the hypotenuse. This is how these problems can be solved:

(a) find the hypotenuse, ‘c’, given legs ‘a’ and ‘b’ of a right triangle.

Given: c2 = a2 + b2,

so: 22 ba +=c For example, if 'a' = 3 and 'b' = 4, then c2 = 32 + 42 = 9 + 16 = 25 so c = 25± = 5 (Rejecting negative value of ‘c’.)

(b) find leg, ‘b’, given leg ‘a’ and hypotenuse ‘c’ of a right triangle.

25

a = 3

b = 4 c = = 5

a = 6

b = 64 = 8

c = 10

Given: c2 = a2 + b2, so: b2 = c2 - a2

so: 22 ac −=b For example, if 'a' = 6 and 'c' = 10, then b2 = 102 - 62 = 100 - 36 = 64 so b = 64± = 8 (Rejecting negative value of ‘b’.)

Do steps 1, 2 and 3 below to practice using the Maths Helper Plus Pythagoras calculator:

1) Start Maths Helper Plus and load the ‘R2 - Pythagoras 2.mhp’ document. This document is a ‘Pythagoras calculator’, which displays the working steps and a labelled diagram for solving many right angled triangle problems. 2) Press the F5 key to display the parameters box: These edit boxes: ‘A’, ‘B’ and ‘C’ are used to enter known values of ‘a’, ‘b’ and ‘c’:

horizontal leg ‘a’

vertical leg ‘b’

hypotenuse ‘c’

3) Enter the two known side lengths, and set the unknown side length to zero. Step 1: Click on the parameters box, on the right side of the value to be changed. Step 2: Press the ‘Backspace’ key to delete the old value. Step 3: Type the new value. Step 4: Click the ‘Update’ button.

click here

NOTE: If the diagram becomes too big for your computer screen, press the ‘F10’ key to make it smaller. To make the diagram bigger, hold down ‘Shift’ while you press ‘F10’. 4) Solve these right angled triangles for the unknown sides. For each triangle:

1. Write down the rule of Pythagoras, 2. Substitute the two known values, 3. Calculate the unknown side length.

1. Enter the given side lengths into the Maths Helper Plus edit boxes. 2. Set the unknown side length to zero.

Check Correct 1. Compare your working and answers with Maths Helper Plus. 2. Fix your mistakes.

Calculate 5) More right angled triangle problems

x

7 24

c

12

(e)

100

(c)

35 37

(d)

x x

60

13 5

x

(a)

(b) 9

Each of the following questions can be solved by applying the rule of Pythagoras.

For each question:

Identify Identify the right angled triangle on your diagram and mark the right angle with a small box:

Label Write the known side lengths on your diagram, then use a letter, such as ‘ x ’, for the unknown side.

Draw Draw a neat diagram of the

situation in the question. Then follow the same steps as in question 4: Calculate, Check and Correct. (a) A trail bike rider travels 21 km due north, then 28 km due east. How far is she from her starting point ? (b) A school science class attached a helium filled balloon to 61 metres of light fishing line. With the free end of the line tied to a tent peg on the school oval, the balloon was released. Eventually the balloon came to rest directly above a point on the oval 11 metres from the tent peg. Assuming the fishing line is pulled straight, calculate the height of the balloon above the ground. (c) Advertised television screen sizes are the diagonal measurement to make them appear more impressive. On a television advertised as having a 51cm screen, the screen height is only 35 cm. What is the screen width ? ( Round your answer to the nearest centimetre. ) (d) In a military exercise, a submarine fired a long range torpedo. After firing, the submarine continued due west at a speed of 16 km / hr, while the torpedo travelled due south at 30 km / hr. How far was the torpedo from the submarine 30 minutes after firing ?

Name: Class: p 19


Trigonometry - Activity 9 Tangent ratio: Introduction.

Consider this right angled triangle:

q

Q

P

side opposite to angle ‘P’

p

side adjacent to angle ‘P’

Side ‘p’ is opposite to angle ‘P’, and side ‘q’ is adjacent to angle ‘P’. (‘Adjacent’ means ‘beside’). In right angled triangles, the term ‘adjacent’ is never used to refer to the longest side opposite to the right angle. This is always called the ‘hypotenuse’. Only the shorter sides, the ‘legs’ of the right triangle, can be ‘adjacent’ sides. 1) What side is opposite angle ‘Q’ ? __________ 2) What side is adjacent to angle ‘Q’ ? ___________

For an angle less than 90º in a right angled triangle, the following ratio: is called the ‘tangent ratio’ of the angle. For example, the tangent ratio of angle ‘P’ above can be found as follows:

qpadjacentoppositePtan

=

=

3) Complete the following:

tan Q =

adjacentopposite

‘tangent ratio of P’ is written as: ‘tan P’

4) Start Maths Helper Plus and load the ‘R2 - Tangent ratio 1.mhp’ document. This document calculates tangent ratios for right triangles. 5) Press the F5 key to display the parameters box: These edit boxes: ‘A’, ‘B’ and ‘C’ are used to set up the right triangle as follows:

vertical leg = A×C

horizontal leg = B×C

multiplier = C

slider

If ‘C’ = 1, then the vertical leg will be ‘A’ units long, and the horizontal leg will be ‘B’ units long. If ‘X’ = 1, then the tangent ratio is calculated for angle ‘P’. If ‘X’ = 2, then the calculations are based on angle ‘Q’. To change a value on the parameters box: (1) click in the centre of its edit box, (2) press the backspace key to delete the old value, (3) type the new value and then (4) click the ‘Update’ button.

6) Complete the tangent ratio calculations for the two marked angles in the right angled triangle below:

(a) (b) oppositeadjacentoppositeQtan =

=

=

=

53adjacent

PtanQ

P 5

3 = =

7) Use Maths Helper Plus to check and correct your work.

• Enter the parameters box values to create the triangles in question 6: Part (a): A = 3, B = 5, C = 1, X = 1 Part (b): A = 3, B = 5, C = 1, X = 2

• Compare the calculations displayed by the computer with yours and correct any mistakes. Hint: Maths Helper Plus always draws the triangle in the same orientation and with the angles named ‘P’ and ‘Q’. This doesn’t matter. The important thing is to enter the opposite and adjacent sides correctly. If ‘X’ = 1 on the parameters box, then ‘A’ is the opposite side and ‘B’ is the adjacent side. If ‘X’ = 2, then ‘B’ is the opposite side and ‘A’ is the adjacent side. 8) Calculate the tangent ratios for each of the two marked angles in the triangles below.

Show all working, and use Maths Helper Plus to help you check and correct your work.

(a) (b)

6

K

J

(b) 2

Y

12

X 5

Investigation: If the size of a triangle is changed, what effect will this have on the tangent ratios? In this activity, you will use Maths Helper Plus to experiment with right angled triangles while recording some results in a table. You will then be able to answer this question and give a reason for your answer. 9) Set up the parameters box as follows: A = 3, B = 4, C = 1, X = 1 The triangle displayed in Maths Helper Plus now has angle ‘P’ = 36.8699º, with the opposite side = 3, adjacent side = 4, and the calculated value of tan 36.8699º = 0.75.

10) Change the size of the triangle in small steps, recording the side lengths and ‘tan P’ values each time. (a) Click on the ‘C’ edit box on the parameters box. (b) Click on the ‘slider’ button. (On the right-hand side of the parameters box.) (c) Quickly press and release the keyboard up arrow key to increase the triangle size. (d) Record the opposite and adjacent side lengths, and ‘tan P’ in the table below.

Repeat these steps until the table is full.

angle P opposite side adjacent side tan P 36.8699º 3 4 0.75

11) Does the size of a triangle have any effect on the tangent ratios of its angles ? Explain your answer:

12) To test your findings from question 11 for some other triangles, change the ‘A’ and ‘C’ values in the parameters box to create other tangent ratios, then vary the ‘C’ value as in ‘10’ above.

p 21 Name: Class:


Trigonometry - Activity 10 Tangent ratio: Finding unknown angles in right triangle.

adjacentopposite

The tangent ratio of an angle θ in a right angled triangle is defined as: tan θ = The value of ‘tan θ’ is found to depend only on the size of the angle, ‘θ’, so that ‘tan 30º’ will be the same for all right angled triangles having a 30º angle. You will now create a table of tangent ratios that can be used to find unknown sides and angles in many different triangles. 1) Start Maths Helper Plus and load the ‘R2 - Tangent ratio 2.mhp’ document. This document calculates tangent ratios for angles in right angled triangles. 2) Press the F5 key to display the parameters box: Edit boxes: ‘A’ and ‘B’ are used as follows:

angle in degrees

side adjacent to the angle

Update button

slider

3) To calculate the tangent ratio for any angle,

• click on the centre of the edit box for ‘A’ on the parameters box. • press backspace to delete the existing angle. • type the new angle. • click the ‘Update’ button.

Use Maths Helper Plus to calculate these tangent ratios: (a) tan 15º = ________ (b) tan 30º = ________ (c) tan 45º = ________ (d) tan 60º = ________ 4) Creating a table of tangent ratios A table of tangent ratios can be used to find unknown angles and sides in right angled triangles. Use Maths Helper Plus to calculate the tangent ratio for the angles in the table below. Write the values in the table. (In the table, the angle is called ‘A’.) Hint: To quickly change the angle on the parameters box, first click on the ‘A’ edit box, backspace and change the number to ‘5’, then click on the slider. Now use the up and down keyboard arrows to change the angle.

Aº tan Aº Aº tan Aº Aº tan Aº

5 35 65

10 40 70

15 45 75

20 50 80

25 55 85

30 60 90 ��

��

��

5) Maths Helper Plus tries to calculate tan 90º, but doesn’t really give the true answer. What is the value of tan 90º ? _________ [ Hint: When ‘A’ = 90º, how long is the opposite side? ]

6) Use the table from question 4 to find the unknown angles below: (a) tan A = 0.2679 (b) tan B = 11.4301 (c) tan C = 1 (d) tan D = 1.7321 (e) tan E = 0.3640 A = _____ B = _____ C = _____ D = _____ E = _____ 7) Find the unknown angles in the right angled triangles below. Use the two sides given to calculate the tangent ratio for the unknown angle, then use the table of tangent ratios that you created in question 4 to find the angle. (a) (b) (c)

25

25 B

tan B = _______

so B = _______ C

3.204

6.871

tan A = _______

so A = _______

A 4.9248

2.84333

(d)

tan C = _______

so C = _______

D 1.02362

11.7

tan D = _______

so D = _______

(e)

E

4.16

11.4295

tan E = _______

so E = _______

Name: Class: p 23


Trigonometry - Activity 11 Tangent ratio: Finding unknown opposite or adjacent sides in right triangles.

In this activity you will practice finding unknown sides of right triangles using the tangent ratio, then use Maths Helper Plus to correct your working and answers. The tangent ratio can be used to calculate the length of one of the shorter sides (legs) of a right triangle. To do this you need the tangent ratio for one of the smaller angles, and the length of one leg of the triangle. There are two types of these problems, depending on whether you are finding the ‘opposite’ or ‘adjacent’ side. Problem type 1: Finding the opposite side. Example 1: Use the tangent ratio to find the unknown side in this triangle:

35º

Solution:

x

°×=

==°

35tan1010

35tan

xso

xadjacentopposite

10 But tan35º = 0.700208 (From a table of tangent ratios, or scientific calculator with a ‘tan’ button.) so x = 10 × 0.700208 = 7.00208

Problem type 2: Finding the adjacent side. Example 2: Use the tangent ratio to find the unknown side in this triangle:

40º

7 Solution:

°=

=×°

==°

40tan7

740tan

740tan

xand

xsoxadjacent

opposite

x

But tan40º = 0.8391 (From a table of tangent ratios, or scientific calculator with a ‘tan’ button.)

34228.88391.07

=

=xso

1) Start Maths Helper Plus and load the ‘R2 - Tangent ratio 3.mhp’ document. This document solves for unknown sides in right triangles using the ‘tan’ ratio. 2) Press the F5 key to display the parameters box. (See below.) ‘A’, ‘B’ and ‘C’ have meanings as shown

below: ‘A’, one of the two smaller angles of the triangle:

The adjacent side to angle ‘A’:

The opposite side to angle ‘A’:

Only one of the legs of the triangle can be entered at a time. Set the unknown side length to zero. To enter a value, click on an edit box, backspace to clear the old value, type the new value then click ‘Update’. The diagram will always draw the adjacent side horizontally, and the opposite side vertically. NOTE: If the diagram becomes too big for your computer screen, press the ‘F10’ key to make it smaller. To make the diagram bigger, hold down ‘Shift’ while you press ‘F10’.

3) Use the tangent ratio to find the unknown leg in each of the triangles below. For each triangle:

Calculate

1. Enter the angle and given side into the Maths Helper Plus edit boxes.

2. Set the unknown side to zero.

1. Identify the opposite and adjacent legs for the angle given.

2. Obtain the tangent ratio of the angle from a calculator or printed table.

3. Write the tangent ratio rule and substitute the known values.

4. Calculate the answer, showing all working steps.

Check Correct

1. Compare your working steps and answers with Maths Helper Plus.

2. Fix your mistakes.

(b) x

14 19º

(a)

30º

x

8.5

25

(d)

x

42º

(c) x

36 70º

(e)

x

9 5º

(f)

x

48º

10

Name: Class: p 25


Trigonometry - Activity 12 Tangent ratio: Angles of elevation and depression.

Angles of elevation and depression are measured from the horizontal, like this:

angle of depression

angle of elevation

observer horizontal

For an object that is higher than the observer, the angle of elevation is measured upwards from the horizontal to the straight line between the observer and the object. For an object that is lower than the observer, the angle of depression is measured downwards from the horizontal to the straight line between the observer and the object. An angle of elevation and depression can help us find an unknown distance. All we need to do is construct a right triangle containing the angle and then use the tan ratio. Case 1 - Angles of elevation Given an angle of elevation and a horizontal or vertical distance between the object and the observer, we can construct a right angled triangle like this:

angle of elevation horizontal

observer

Now we can use the tan ratio to find the unknown horizontal or vertical side. Case 2 - Angles of depression Given an angle of depression and a horizontal or vertical distance between the object and the observer, we can construct a right angled triangle like this:

angle of

A

A

depression

horizontal

observer

The angle of depression, ‘A’, is the same as the angle in the triangle (near the dog) because the base of the triangle is parallel to the horizontal. Once again, the tan ratio can be used to find the unknown horizontal or vertical side of this triangle.

1) Start Maths Helper Plus and load the ‘R2 - Tangent ratio 4.mhp’ document. This document solves ‘angle of elevation’ and ‘angle of depression’ problems. 2) Press the F5 key to display the parameters box. (See below.) ‘A’, ‘B’, ‘C’ and ‘X’ have meanings as shown:

Vertical side ( ):Set to zero if unknown

Horizontal side ( ): Set to zero if unknown

Angle of elevation or depression: Set ‘X’ to: ‘ ’ if ‘A’ is an angle of , 1 elevation or ‘ ’ if ‘A’ is an angle of . depression2

3) For each of the word problems below:

1. Draw a labelled diagram of the situation similar to those on the front of this sheet. Include the horizontal line of sight from the observer and a right angled triangle. 2. Use the tangent ratio to calculate the unknown leg of the right triangle. Show all working. 3. Make sure you have answered the original question. Include units if necessary. 4. Use Maths Helper Plus to correct your diagram, working steps and answer.

NOTE: If the diagram becomes too big for your computer screen, press the ‘F10’ key to make it smaller. To make the diagram bigger, hold down ‘Shift’ while you press ‘F10’. a) A ladder placed on a flat horizontal surface rests against a vertical wall with an angle of elevation of 65º. The foot of the latter is 2 metres from the base of the wall. Find the hight of the point where the ladder touches the wall. To check with Maths Helper Plus, set: A=65, B=2, C=0, X=1 b) A hungry cat spies a bird in a tree. The bird is 3.4 metres above the ground, and the angle of elevation to the bird from the cat’s point of view is 40º. How far is the cat from the base of the tree ? To check with Maths Helper Plus, set: A=40, B=0, C=3.4, X=1 c) The common archer fish (Toxotes chatareus) is able to squirt a water jet from its mouth with deadly accuracy to shoot down insects crawling on leaves and stems of overhanging vegetation. An archerfish at the surface of the water notices a dragonfly resting on a twig at an angle of elevation 77º. The dragonfly is exactly above a point on the water that is 30cm from the fish. How high is the dragonfly above the water surface ? To check with Maths Helper Plus, set: A=77, B=30, C=0, X=1 d) A parachutist has an altitude of 320 metres when directly above a high voltage power pole. She spies the landing target and estimates that the angle of depression to the target is 33º. What is the approximate horizontal distance from the power pole to the target ? To check with Maths Helper Plus, set: A=33, B=0, C=320, X=2 e) A fishing boat is 120 metres away from a point on the water that is directly above a school of large fish. The sonar fish detector on the boat indicates that the angle of depression from the boat to the fish is 55º. How deep are the fish ? To check with Maths Helper Plus, set: A=55, B=120, C=0, X=2

Name: Class: p 27


Trigonometry - Activity 13 Sine ratio: Introduction.

Consider this right angled triangle:

q

Q

P

side opposite to angle ‘P’

p

side opposite to angle ‘Q’

Side ‘p’ is opposite to angle ‘P’, and side ‘q’ is opposite to angle ‘Q’.

For an angle less than 90º in a right angled triangle, the following ratio: is called the ‘sine ratio’ of the angle. hypotenuse

opposite

For example, the sine ratio of angle ‘P’ above can be found as follows:

hphypotenuse

oppositePsin

=

=

‘sine ratio of P’ is written as: ‘sin P’


sin Q = 2) Start Maths Helper Plus and load the ‘R2 - Sine ratio 1.mhp’ document. This document calculates sine ratios for right triangles. 3) Press the F5 key to display the parameters box: These edit boxes: ‘A’, ‘B’, ‘C’ and ‘X’ are used to set up the right triangle as follows:

opposite side = A×C

hypotenuse = B×C

multiplier = C

slider button

If ‘C’ = 1, then the vertical leg will be ‘A’ units long, and the hypotenuse will be ‘B’ units long. If ‘X’ = 1, then the sine ratio is calculated for angle ‘P’, and the opposite side A×C is vertical. If ‘X’ = 2, then the sine ratio is calculated for angle ‘Q’, and the opposite side A×C is horizontal. To change a value on the parameters box: (1) click in the centre of an edit box, (2) press the backspace key to delete the old value, (3) type the new value and then (4) click the ‘Update’ button.

4) Complete the sine ratio calculations for the two marked angles in the right angled triangle below:

(a) (b) oppositehypotenuse

oppositeQsin =

=

=

=

106hypotenuse

Psin Q

P

10 6

= =

8 5) Use Maths Helper Plus to check and correct your work.


• Compare the calculations displayed by the computer with yours and correct any mistakes. 6) Calculate the sine ratios for the two marked angles in these triangles.

Show all working, and use Maths Helper Plus to help you check and correct your work. Hint: When correcting your work, if the diagram is too big in Maths Helper Plus, then press the F10 key to make it smaller. If it is too small, then hold down a Shift key and press F10 to make it bigger.

(a) (b)

5.65685

P

Q

15

N

M

10 4

11.1803 4

(c) (d)

19.2094

K

J

(b) 12.6

Y

3.2

13

X 15 12

Investigation: If the size of a triangle is changed, what effect will this have on the sine ratios ?

Set the parameters box values to: A = 3, B = 4, C = 1, X = 1 The triangle displayed in Maths Helper Plus now has angle ‘P’ = 48.5904º, opposite side 3, hypotenuse 5, and calculated value of sin 48.5904º = 0.75.

Gradually change the scale of the triangle measurements, as follows: (a) Click on the ‘C’ edit box on the parameters box. (b) Click on the ‘slider’ button. (c) Repeatedly press the keyboard up arrow key to increase the triangle size. (d) Repeatedly press the keyboard down arrow key to decrease the triangle size.

As the triangle changes size, watch the values of angle ‘P’ and the sine ratio on the computer screen. 7) Does the size of a triangle have any effect on the sine ratio of its angles ? Explain.

Name: Class: p 29


Trigonometry - Activity 14 Sine ratio: Finding unknown angles in right triangles.

hyotenuseopposite

The sine ratio of an angle θ in a right angled triangle is defined as: sin θ = The value of ‘sin θ’ is found to depend only on the size of the angle, ‘θ’, so that ‘sin 30º’ will be the same for all right angled triangles having a 30º angle. You will now create a table of sine ratios that can be used to find unknown sides and angles in many different triangles. 1) Start Maths Helper Plus and load the ‘Trig1 - Sine ratio 2.mhp’ document. This document calculates sine ratios for angles in right angled triangles. 2) Press the F5 key to display the parameters box: Edit boxes: ‘A’ and ‘B’ are used as follows:

angle in degrees


Update button

slider

3) To calculate the sine ratio for any angle,


Use Maths Helper Plus to calculate these sine ratios: (a) sin 15º = ________ (b) sin 30º = ________ (c) sin 45º = ________ (d) sin 60º = ________ 4) Creating a table of sine ratios A table of sine ratios can be used to find unknown angles and sides in right angled triangles. Use Maths Helper Plus to calculate the sine ratio for the angles in the table below. Write the values in the table. (In the table, the angle is called ‘A’.) Hint: To quickly change the angle on the parameters box, first click on the ‘A’ edit box, backspace and change the number to ‘5’, then click on the slider. Now use the up and down keyboard arrows to change the angle.

Aº sin Aº Aº sin Aº Aº sin Aº

5 35 65

10 40 70

15 45 75

20 50 80

25 55 85

30 60 90 5) The sine ratio can never be greater than 1. Why is this ?

6) Use the table from question 4 to find the unknown angles below:

(a) sin A = 0.3746 (b) sin B = 0.8192 (c) sin C = 1 (d) sin D = 23

(e) sin E = 21

A = _____ B = _____ C = _____ D = _____ E = _____ 7) Find the unknown angles in the right angled triangles below. Use the two sides given to calculate the sine ratio for the unknown angle, then use the table of sine ratios that you created in question 4 to find the angle. (a) (b) (c) (d)

3.50104 6.10387

A

D

sin A = _______

so A = _______ so B = _______

21.7801 16.6846

B

8.77141

8.24243

15 30

(e)

C

22.6851

23.0351 E

sin B = _______

sin C = _______

so C = _______

so D = _______

sin D = _______

sin E = _______

so E = _______

Name: Class:

p 31 Maths Helper Plus Resource Set 2. Copyright © 2003 Bruce A. Vaughan, Teachers’ Choice Software

Trigonometry - Activity 15 Sine ratio: Finding the unknown opposite side or hypotenuse in right triangles.

In this activity you will practice finding an unknown opposite side or hypotenuse of right triangles using the sine ratio, then you will use Maths Helper Plus to correct your working and answers. The sine ratio can be used to calculate the length of one of the shorter sides (legs) of a right triangle, or to find the length of the hypotenuse. In both cases you need to know the sine ratio for one of the smaller angles. If you are finding an unknown leg length, then the hypotenuse must be known. If you are finding the hypotenuse, then one leg of the triangle must be known. The steps for both problem types are shown below: Problem type 1: Finding the opposite side. Example 1: Use the sine ratio to find the unknown opposite side ‘x’ in this triangle:

10

30º

Solution:

x

°×=

==°

30sin1010

30sin

xso

xhypotenuse

opposite

But sin30º = 0.5 (From a table of sine ratios, or scientific calculator with a ‘sin’ button.) so x = 10 × 0.5 = 5

x

40º

Problem type 2: Finding the hypotenuse. Example 2: Use the sine ratio to find the hypotenuse in this triangle: 9 Solution:

°=

=×°

==°

40sin9

940sin

940sin

xand

xsoxhypotenuse

opposite

But sin 40º = 0.642788 (From a table of sine ratios, or scientific calculator with a ‘sin’ button.)

0015.14642788.0

9

=

=xso

1) Start Maths Helper Plus and load the ‘Trig1 - Sine ratio 3.mhp’ document. This document solves for unknown sides in right triangles using the ‘sine’ ratio. 2) Press the F5 key to display the parameters box. (See below.) ‘A’, ‘B’ and ‘C’ have meanings as shown below: ‘A’, one of the two smaller angles of the triangle:

The opposite side to angle ‘A’:

The hypotenuse:

Only one side length can be entered at a time. Set the unknown side length to zero. To enter a value, click on an edit box, backspace to clear the old value, type the new value then click ‘Update’. NOTE: If the diagram becomes too big for your computer screen, press the ‘F10’ key to make it smaller. To make the diagram bigger, hold down ‘Shift’ while you press ‘F10’.

3) Use the sine ratio to find the unknown side ‘x’ in each of the triangles below. For each triangle:

75º

x

45

(b)

x

22

40º

1. your working steps and answers with Maths Helper Plus.

2. your mistakes. Fix

Compare

Check Correct Calculate

1. the angle and given side into the Maths Helper Plus edit boxes.

2. the unknown side to zero. Set

Enter

1. the opposite side for the angle given, and the hypotenuse.

2. of the angle from a calculator or printed table.

3. Write the sine ratio rule and the known values. substitute

4. the answer, showing all working steps. Calculate

Obtain the sine ratio

Identify

(e) Challenge question!

50º

10 25

35º

x

x

x

10 25º

(a)

30º

8 x

(c)

(d)

(f) Challenge question!

Name: Class:

p 33 Maths Helper Plus Resource Set 2. Copyright © 2003 Bruce A. Vaughan, Teachers’ Choice Software

Trigonometry - Activity 16 Cosine ratio: Introduction.

Consider this right angled triangle: Q

side adjacent to angle ‘Q’

p P

side adjacent to angle ‘P’

q Side ‘p’ is adjacent to angle ‘Q’, and side ‘q’ is adjacent to angle ‘P’. (‘Adjacent’ means ‘beside’)

For an angle less than 90º in a right angled triangle, the following ratio: is called the ‘cosine ratio’ of the angle. hypotenuse

adjacent

For example, the cosine ratio of angle ‘P’ above can be found as follows:

hqhypotenuse

adjacentP cos

=

=

‘cosine ratio of P’ is written as: ‘cos P’


cos Q = 2) Start Maths Helper Plus and load the ‘R2 - Cosine ratio 1.mhp’ document. This document calculates cosine ratios for right triangles. 3) Press the F5 key to display the parameters box: These edit boxes: ‘A’, ‘B’, ‘C’ and ‘X’ are used to set up the right triangle as follows:

adjacent side = A×C

slider button

hypotenuse = B×C

multiplier = C If ‘C’ = 1, then the vertical leg will be ‘A’ units long, and the hypotenuse will be ‘B’ units long. If ‘X’ = 1, then the cosine ratio is calculated for angle ‘P’, and the adjacent side A×C is horizontal. If ‘X’ = 2, then the cosine ratio is calculated for angle ‘Q’, and the adjacent side A×C is vertical. To change a value on the parameters box: (1) click in the centre of an edit box, (2) press the backspace key to delete the old value, (3) type the new value and then (4) click the ‘Update’ button.

4) Complete the cosine ratio calculations for the two marked angles in the right angled triangle below:

(a) (b) adjacenthypotenuse

adjacentQ cos =

=

=

=

1512hypotenuse

P cosQ

15 9

= P =

12 5) Use Maths Helper Plus to check and correct your work.


• Compare the calculations displayed by the computer with yours and correct any mistakes. 6) Calculate the cosine ratios for the two marked angles in these triangles.

Show all working, and use Maths Helper Plus to help you check and correct your work. Hint: When correcting your work, if the diagram is too big in Maths Helper Plus, then press the F10 key to make it smaller. If it is too small, then hold down a Shift key and press F10 to make it bigger.

Investigation: If the size of a triangle is changed, what effect will this have on the cosine ratios ?

Set the parameters box values to: A = 3, B = 5, C = 1, X = 1 The triangle displayed in Maths Helper Plus now has angle ‘P’ = 53.1301º, side adjacent to angle ‘P’ = 3, hypotenuse = 5, and calculated value of cos 53.1301º = 0.6

Gradually change the scale of the triangle measurements, as follows: (a) Click on the ‘C’ edit box on the parameters box. (b) Click on the ‘slider’ button. (c) Repeatedly press the keyboard up arrow key to increase the triangle size. (d) Repeatedly press the keyboard down arrow key to decrease the triangle size.

As the triangle changes size, watch the values of angle ‘P’ and the cosine ratio on the computer screen. 7) Does the size of a triangle have any effect on the cosine ratio of its angles ? Explain.

M

N

13 5

12

Q

P

13.4536

9 (c)

44

Y

9.43398

45

X

(b)

(a)

9

15 J (d)

30 25.9808

K

Name: Class: p 35


Trigonometry - Activity 17 Cosine ratio: Finding unknown angles in right triangles.

hyotenuseadjacent

The cosine ratio of an angle θ in a right angled triangle is defined as: cos θ = The value of ‘cos θ’ is found to depend only on the size of the angle, ‘θ’, so that ‘cos 30º’ will be the same for all right angled triangles having a 30º angle. You will now create a table of cosine ratios that can be used to find unknown sides and angles in many different triangles. 1) Start Maths Helper Plus and load the ‘R2 - Cosine ratio 2.mhp’ document. This document calculates cosine ratios for angles in right angled triangles. 2) Press the F5 key to display the parameters box: Edit boxes: ‘A’ and ‘B’ are used as follows:

angle in degrees


Update button

slider

3) To calculate the cosine ratio for any angle,


Use Maths Helper Plus to calculate these cosine ratios: (a) cos 10º = ________ (b) cos 30º = ________ (c) cos 45º = ________ (d) cos 60º = ________ 4) Creating a table of cosine ratios A table of cosine ratios can be used to find unknown angles and sides in right angled triangles. Use Maths Helper Plus to calculate the cosine ratio for the angles in the table below. Write the values in the table. (In the table, the angle is called ‘A’.) Hint: To quickly change the angle on the parameters box, first click on the ‘A’ edit box, backspace and change the number to ‘5’, then click on the slider. Now use the up and down keyboard arrows to change the angle.

Aº cos Aº Aº cos Aº Aº cos Aº

5 35 65

10 40 70

15 45 75

20 50 80

25 55 85

30 60 90 5) The cosine ratio can never be greater than 1. Why is this ?

6) Use the table from question 4 to find the unknown angles below:

(a) cos A = 1 (b) cos B = 21

(c) cos C = 0.5736 (d) cos D = 23

(e) cos E = 0.7071

A = _____ B = _____ C = _____ D = _____ E = _____ 7) Find the unknown angles in the right angled triangles below. Use the two sides given to calculate the cosine ratio for the unknown angle, then use the table of cosine ratios that you created in question 4 to find the angle.

cos B = _______ (a) (b)

so B = _______ cos A = _______ so A = _______ (c)

cos C = _______

so C = _______

cos E = _______

so E = _______

A 12.66

6.33

4.24264 3

(e)

(d)

7.77862

so D = _______

D 5

10.987

E

cos D = _______

B

C

13.200513

9

Name: Class:

Trigonometry - Activity 18 Cosine ratio: Finding the unknown adjacent side or hypotenuse in right triangles.

In this activity you will practice finding an unknown opposite side or hypotenuse of right triangles using the cosine ratio, then you will use Maths Helper Plus to correct your working and answers. The cosine ratio can be used to calculate the length of one of the shorter sides (legs) of a right triangle, or to find the length of the hypotenuse. In both cases you need to know the cosine ratio for one of the smaller angles. If you are finding an unknown leg length, then the hypotenuse must be known. If you are finding the hypotenuse, then one leg of the triangle must be known. The steps for both problem types are shown below: Problem type 1: Finding the adjacent side. Example 1: Use the cosine ratio to find the unknown adjacent side ‘x’ in this triangle: Solution:

°×=

==°

60cos2020

60cos

xso

xhypotenuseadjacent

But cos 60º = 0.5 (From a table of cosine ratios, or scientific calculator with a ‘cos’ button.) so x = 20 × 0.5 = 10

Problem type 2: Finding the hypotenuse. Example 2: Use the cosine ratio to find the hypotenuse in this triangle: Solution:

°=

=×°

==°

35cos8

835cos

835cos

xand

xsoxhypotenuse

adjacent

p 37

60º

20

x

35º

x


8

But cos 35º = 0.8192 (From a table of cosine ratios, or scientific calculator with a ‘cos’ button.)

766.98192.08

=

=xso

1) Start Maths Helper Plus and load the ‘Trig1 - Cosine ratio 3.mhp’ document. This document solves for unknown sides in right triangles using the ‘cosine’ ratio. 2) Press the F5 key to display the parameters box. (See below.) ‘A’, ‘B’ and ‘C’ have meanings as shown below: ‘A’, one of the two smaller angles of the triangle:

The adjacent side to angle ‘A’:

The hypotenuse:

Only one side length can be entered at a time. Set the unknown side length to zero. To enter a value, click on an edit box, backspace to clear the old value, type the new value then click ‘Update’. NOTE: If the diagram becomes too big for your computer screen, press the ‘F10’ key to make it smaller. To make the diagram bigger, hold down ‘Shift’ while you press ‘F10’.

3) Use the cosine ratio to find the unknown side ‘x’ in each of the triangles below. For each triangle:

Check Correct Calculate 1. the adjacent side for the

angle given, and the hypotenuse. Identify

1. the angle and given side into the Maths Helper Plus edit boxes.

2. the unknown side to zero. Set

Enter 1. your working steps and answers with Maths Helper Plus.

2. your mistakes. Fix

Compare2. of the angle from a calculator or printed table. Obtain the cosine ratio

3. Write the cosine ratio rule and the known values. substitute

4. the answer, showing all working steps. Calculate

10

x 30º

(a)

(b)

(c)

(d)

(e) Challenge question!

(f) Challenge question!

8.82702

25º

65º

12

x

x

20

x

x 33.1013 14

x

85º

45º

70º

Name: Class: p 39


Trigonometry - Activity 19 Similar triangles: Definition, Equivalence of corresponding angles, Scale factor.

‘Similar’ triangles have the same shape. For example, these two triangles are similar:

90º

53.1301º

3

5

36.8699º

90º

53.1301º 36.8699º

4 4.5 6

7.5

Triangle 1 Triangle 2

Scale drawings of any shape are said to be ‘similar’ to one another. Features such as points, lines and angles found on one shape can be identified on any other similar shape. Features on one shape are said to correspond to features on a shape similar to it. The diagrams above show two triangles that are similar. Side ‘5’ is the hypotenuse of triangle 1 . Because the triangles are similar, the hypotenuse in triangle 2 (side 7.5) corresponds to the hypotenuse in triangle 1 (side 5). The following two properties are always true for a pair of similar triangles.

a. Corresponding angles are equal. b. The ratio of corresponding sides is a constant. This constant is called the ‘scale factor’.

The tables below contain information about the two similar triangles above. Complete the missing entries...

1) Corresponding angles: Write the two unknown angles in the table below...

Angle in triangle 1 Corresponding angle in triangle 2

90º 90º

53.1301º

36.8699º

2) Ratio of corresponding sides: Calculate the scale factor for the last two rows of this table. Write your answers in the table...

s1 = Side in triangle 1 s2 = Corresponding side in triangle 2 Scale factor: 1

2

ssr =

3 4.5 5.135.4

==r

4 6

5 7.5

You will now use Maths Helper Plus to correct these answers and to demonstrate these properties of similar triangles. 3) Start Maths Helper Plus and load the ‘R2 - Similarity 1.mhp’ document. This document displays two similar triangles and some calculations.

4) Press the F5 key to display the parameters box. (See below.) edit boxes slider The variables in the edit boxes have these meanings: Update button

• ‘A’, ‘B’ and ‘C’ are the side lengths of triangle 1

• ‘X’ is the ratio of corresponding sides between triangle 1 and triangle 2.

• ‘D’ shifts triangle 2 to the right. If ‘D’ = 0, then the two similar triangles lie on top of each other. To type a new value for A, B, C, D, or X, you: click on the edit box, use the ‘Backspace’ key to delete the old value, then type the new value and click the Update button. 5) Correct your answers to questions (1) and (2) above.

(a) Set the variables in the parameters box to be as follows: A = 3, B = 4, C = 5, D = 6, X = 1.5

(b) The sides in the shaded triangle are now 1.5 times larger than the other triangle. Are the side lengths of the shaded triangle the same as for ‘triangle 2’ on the front of this sheet?

NOTE: If the diagram becomes too big for your computer screen, press the ‘F10’ key to make it smaller. To make the diagram bigger, hold down ‘Shift’ while you press ‘F10’. 6) Experimenting with the scale factor

• Click on the ‘X’ edit box, change the value to 1, then click on the slider button. • Use the up and down keyboard arrow keys to increase and decrease the scale factor.

7) What effect does the scale factor have on the shaded triangle’s size when it is:

(a) greater than 1

(b) equal to 1

(c) less than 1 8) How does the scale factor effect the angles in the shaded triangle ? 9) Experimenting with the triangle shape

• Set the scale factor ‘X’ to 1.5

• Click on the ‘A’ edit box.

• Click on the slider button with the mouse, then drag it up and down slowly. This will change the length of side ‘A’ in triangle 1.

(a) Do the two similar triangles always have the same shape ? (b) Stop dragging the slider button at a few different positions. At each position, compare the pairs of corresponding angles in the two similar triangles. Are the angles in one triangle always the same as the corresponding angles in the other triangle ?

(c) Use the slider to change the other side lengths, ‘B’ and ‘C’. In each case, compare the corresponding angles as you did in (b) above.

Name: Class: p 41


Trigonometry - Activity 20 Similar triangles: Applications.

If two triangles are known to be similar, then corresponding angles are equal, and the ratio of corresponding sides is a constant. Consider these two similar triangles: If the corresponding angles are equal, what are angles ‘P’, ‘Q’ and ‘R’ ?

P = ___________ Q = ___________ R = ___________

The ratio of corresponding sides is: 25

10= . Side ‘a’ corresponds to length ‘4’. So 2

4=

a, and ‘a’ = 8.

1) Calculate the length of side ‘b’.

90º

36.8699º 53.1301º

3

4

5 10

R

Q

P a

b

triangle 2 triangle 1

2) Start Maths Helper Plus and load the ‘R2 - Similarity 1.mhp’ document. This document displays two similar triangles and some calculations. 3) Press the F5 key to display the parameters box. (See below.) edit boxes slider The variables in the edit boxes have these meanings: Update button

• ‘A’, ‘B’ and ‘C’ are the side lengths of triangle 1

• ‘X’ is the ratio of corresponding sides between triangle 1 and triangle 2.

• ‘D’ shifts triangle 2 to the right. If ‘D’ = 0, then the two similar triangles lie on top of each other. To type a new value for A, B, C, D, or X, you: click on the edit box, use the ‘Backspace’ key to delete the old value, then type the new value and click the Update button. 4) Correct your answer to question (1) above.

(a) Set the variables in the parameters box to be as follows: A = 3, B = 4, C = 5, D = 6, X = 2

(b) The sides in the shaded triangle are all 2 times larger than the other triangle. Was your answer for question (1) correct ?

NOTE: If the diagram becomes too big for your computer screen, press the ‘F10’ key to make it smaller. To make the diagram bigger, hold down ‘Shift’ while you press ‘F10’.

5) Six pairs of similar triangles are shown below. For each pair, calculate the ratio of corresponding sides, then find the unknown side lengths.

(a)

A = 5, B = 5.5, C = 5, D = 6

5.5 5

x y 6 10

A = 8, B = 10, C = 6, D = 7

5 x

(b)

y 6 8 5

A = 4, B = 7.5, C = 10, D = 11

y

6

x 7.5

10

4

(c)

x

y 14

33

29 20

A = 4, B = 7, C = 6, D = 0

A = 20, B = 29, C = 33, D = 36

(f)

(e)

4

6 7

2.2

x y

A = 7, B = 13, C = 18, D = 19

y

12

x

7

13

18

(d)

6) Use Maths Helper Plus to check and correct your work. • Enter the parameters box values included with each question above to create the triangles. • Enter the scale factor you calculated in edit box ‘X’. • Click the ‘Update’ button on the parameters box. • Compare your calculations and answers with those displayed by the computer. Correct your mistakes.

Name: Class: p 43


Trigonometry - Activity 21 b

C a

B c

A

General Triangle Solution: Given three sides.

When the three side lengths 'a', 'b' and 'c' of a triangle are known, then the three internal angles 'A', 'B' and 'C' can be found. ( See diagram at right. ) These are the formulas used to solve this type of problem:

The sum of the internal angles equals 180º ...

A + B + C = 180º

The 'sine rule' ...

Cc

Bb

Aa

sinsinsin==

The 'cosine rule' ...

a² = b² + c² − 2bc cosA or

b² = a² + c² − 2ac cosB or

c² = b² + a² − 2ba cosC

We will demonstrate the procedure for solving a triangle given

b = 3 C

a = 2

B A

three sides, using this triangle as an example:

c = 4 The triangle is now solved. This diagram shows all of the sides and angles:

1. Use the cosine rule to find the largest angle. (Which is opposite the largest side.)

2. Use the sine rule to find one of the remaining angles.

3. Use the ‘sum of internal angles’ rule to find the remaining angle.

25.0322432

2cos

cos2

222

222

222

−=××−+

=

−+=

−+=

abcbaC

Cbaabc

Find the inverse cos of -0.25 using a scientific calculator... C = cos−1(−0.25) = 104.478º

484123.04

°478.104sin2

sinsin

sinsin

=

=

=

=

cCaA

Cc

Aa

Find the inverse sin of 0.484123 using a scientific calculator... A = sin−1(0.484123) = 28.955º

The sum of the internal angles equals 180º ... A + B + C = 180º so B = 180º − (A+C) = 180º − (28.955º + 104.478º) = 180º − 133.433º = 46.5675º

NOTE: There can only be one angle in a triangle that is obtuse (greater than 90°). If a triangle has an obtuse angle, then it will be opposite the largest side. The reason for finding it first is that in the next step we will use the sine rule to find the second angle. The inverse sin operation that we will use can only give us acute angles (less than 90°), so we avoid a possible wrong answer by first eliminating the only possibility of an obtuse angle. A = 28.955º

c = 4

B = 46.5675º

a = 2

C = 104.478º

b = 3

1) Six triangles are shown below. For each triangle, find the three interior angles. Show all working steps.

b = 45

a = 45

(c)

C

B A

Use Maths Helper Plus to correct your work after solving each triangle.

b = 3 a = 5

(a) C

A

28 11

c = 4

(d)

C

B A

B

a = 3 b = 5

(b)

C

B A c = 6

c = 15

18

8

5 5

(f)

11

21 5(e)

To correct your work... 2) Start Maths Helper Plus and load the ‘R2 - Triangle_sover_SSS.mhp’ document. This document shows the working steps and a diagram for solving triangles when given three side lengths. 3) Display the triangle solver options box Double click the mouse in the border to the left of the calculations. ( This area is shaded in the diagram below.) The triangle solver options box will display its 'Lengths & Angles' tab...

Click the 'Clear' button to remove the previous triangle, then click on the 'a' edit box. Now type the length for side 'a' of your triangle. Repeat for 'b' and 'c'. Click the 'Apply' button at the bottom of the edit box. The calculated values will display on the options box. Click the 'OK' button to close the options box. The calculations and triangle diagram will be displayed on your screen. NOTE: If the diagram becomes too big for your computer screen, press the ‘F10’ key to make it smaller. To make the diagram bigger, hold down ‘Shift’ while you press ‘F10’.

Name: Class: p 45


Trigonometry - Activity 22 General Triangle Solution: Given two sides and the included angle.

The angle between two sides of a triangle is called the included angle of those sides. If any two sides and the included angle are known for a triangle, then the remaining side length and the two unknown angles can be found. In the diagram (right), the two given sides are b = 3 and c = 4. The included angle between these sides is A = 30°. These are the formulas used to solve this type of problem:


A + B + C = 180º The 'sine rule' ...

Cc

Bb

Aa

sinsinsin==

The 'cosine rule' ...

a² = b² + c² − 2bc cosA or

b² = a² + c² − 2ac cosB or

c² = b² + a² − 2ba cosC

We will demonstrate the procedure for solving a triangle given two sides and the included angle, using the triangle given above. These are the steps: The triangle is now solved. This diagram shows all of the sides and angles:

1. Use the cosine rule to find the unknown side.

2. Use the sine rule to find the smaller of the two unknown angles. ( A smaller angle will be opposite a smaller side. )

3. Use the ‘sum of internal angles’ rule to find the remaining angle.

94.21539030°30cos43243

cos222

222

=××+=

−+= Abccba

Taking the positive square root... a = 2.05314

730588.005314.2

°30sin3

sinsin

sinsin

=

=

=

=

aAbB

Bb

Aa

Find the inverse sin of 0.730588 using a scientific calculator... A = sin−1(0.730588) = 46.9357º

The sum of the internal angles equals 180º ... A + B + C = 180º so C = 180º − (A+B) = 180º − (30º + 46.9357º) = 180º − 76.9357º = 103.064º

A = 30º B = 46.9357º

a = 2.05314

C = 103.064º

b = 3

A = 30° c = 4

B

a C

b = 3

c = 4

1) Six triangles are shown below. For each triangle, find the three interior angles using the method described on the front of this sheet. Show all working steps. Use Maths Helper Plus to correct your work after solving each triangle.

B

C a

A = 55°

b = 3

(a)

c = 18

a = 15

C

B = 30°

A

(c)

15

19

10°

(d)

c

B

b = 9

A

C = 90° (b)

a = 5

c = 6

(f)

50 49

62°

13

13

22°

(e)

To correct your work... 2) Start Maths Helper Plus and load the ‘R2 - Triangle_sover_SAS.mhp’ document. This document shows the working steps and a diagram for solving triangles given two sides and the included angle. 3) Display the triangle solver options box Double click the mouse in the border to the left of the calculations. ( This area is shaded in the diagram below.) The triangle solver options box will display its 'Lengths & Angles' tab (See below) ... Click the 'Clear' button to remove the previous triangle, then enter your given angle and two sides. You can enter these in three ways, but the answer will be correct as long as the given angle is between the given sides. ( In this diagram we are entering sides 'b' and 'c', and the angle 'A' which is between 'b' and 'c'. ) To enter a value, click on its edit box, then type the value.

IMPORTANT Do not use the degree operator ° for angles, eg type 35 degrees as just 35, not as 35°.

Click the 'Apply' button at the bottom of the dialog box. The calculated values will display on the options box. Click the 'OK' button to close the options box. The calculations and triangle diagram will be displayed on your screen. NOTE: If the diagram becomes too big for your computer screen, press the ‘F10’ key to make it smaller. To make the diagram bigger, hold down ‘Shift’ while you press ‘F10’.

Name: Class: p 47


Trigonometry - Activity 23 General Triangle Solution: Given two sides and a non-included angle.

A ‘non-included’ angle is an internal angle of a triangle that does not lie between two given sides. If two sides and a non-included angle are given for a triangle, then there may be up to two different possible triangles that have these measurements. For example, both triangles in the diagram below have sides 7 and 5, and a non-included angle of 30°...

c

B c

a = 5 b = 7

case 1: b = 7 a = 5

C case 2:C

A = 30º A = 30º We will demonstrate the procedure for solving a triangle given two sides and a non-included angle, using the triangle given above. These are the steps:

7.05

°30sin7

sinsin

sinsin

=

=

=

=

aAbB

Bb

Aa

1. Use the sine rule to find the unknown angle that is opposite one of the given sides. Subtract this angle from 180º to find a second angle. Test your calculated angles by comparing them with the given angle. In this case, because side b > side a, angle B must be > angle A. Since both values of B we have found are greater than A = 30º, we can accept them both. If either B value were less than 30º it would mean that this triangle only had one solution.

2. Use the ‘sum of internal angles’ rule tofind the remaining angle.

3. Use the sine rule to find the remaining unknown side.

The sum of the internal angles equals 180º ... A + B + C = 180º so Case 1: C = 180º − (A+B) = 180º − (30º + 44.427º) = 180º − 74.427º = 105.573º

Case 1: Find the inverse sin of 0.730588 using a scientific calculator... B = sin (0.7) = 44.427º −1

Subtract this angle from 180º to find a second value for angle A: Case 2: B = ( 180º - 44.427º ) = 135.573º

Case 2: C = 180º − (A+B) = 180º − (30º + 135.573º) = 180º − 165.573º = 14.427º

63289.9°30sin

°573.105sin5sinsin

sinsin

=

=

=

=

ACac

Cc

Aa

Case 1: Case 2:

49146.2°30sin

°427.14sin5sinsin

sinsin

=

=

=

=

ACac

Cc

Aa

Case 1: Case 2:

7 7

5

5

9.63289 2.49146 30° 30° 44.427°

105.573°

14.427°

These triangles show the case 1 and case 2 solutions...

135.573°

1) In step 1 of the triangle solution on the front of this sheet, a scientific calculator is used to find the inverse sine of 0.7. This is angle B, and = 44.427º. Explain why another value of angle B is given by : ( 180º - 44.427º ) = 135.573º 2) Given two sides and a non-included angle, sometimes there are two possible triangles as in the example on the front of this sheet, but sometimes there may be only 1, or even none at all. Explain what will happen in step 1 of the procedure for each of these situations. 3) Three triangles are shown below. In each case, two sides and a non-included angle are given. For each triangle:

(a) If possible, sketch on the diagrams to show how a second triangle can be constructed with the same three measurements. Shade the second triangle ( if there is one ). (b) Solve the triangle for the unknown side and angles using the method on the front of this sheet.

Draw labelled diagrams of the triangles. To correct your work...

B

b = 9

c

a = 28

B = 15° A

A = 25° B

C

A = 90°

C

c

b = 10 b a = 8 a = 5

c = 5

C

2) Start Maths Helper Plus and load the ‘Triangle_sover_ASS.mhp’ document.

This document shows the working steps and a diagram for solving triangles when given two sides and a non-included angle.

3) Display the triangle solver options box Double click the mouse in the border to the left of the calculations. ( This area is shaded in the diagram below.) The triangle solver options box will display its 'Lengths & Angles' tab...

IMPORTANT Do not use the degree operator ° for angles, eg type 35 degrees as just 35,

as 35°. not

Click the 'Clear' button to remove the previous triangle. Enter the sides and angles into three of the white edit boxes. Click the 'Apply' button at the bottom of the edit box. The calculated values will display on the options box. Click the 'OK' button to close the options box. The calculations and triangle diagram will be displayed on your screen. NOTE: If the diagram becomes too big for your computer screen, press the ‘F10’ key to make it smaller. To make the diagram bigger, hold down ‘Shift’ while you press ‘F10’.

Name: Class: p 49


Trigonometry - Activity 24 General Triangle Solution: Given two angles and one side.

When two internal angles and one side of a triangle are known, the other two internal angle and the unknown sides can be found. These are the formulas used to solve this type of problem:


A + B + C = 180º

The 'sine rule' ...

Cc

Bb

Aa

sinsinsin==

We will demonstrate the procedure for solving a triangle given two angles and one side, using the triangle given above. These are the steps:

19204.4°70sin°80sin4

sinsin

sinsin

=

=

=

=

BCbc

Bb

Cc

The sum of the internal angles equals 180º ... A + B + C = 180º so C = 180º − (A+B) = 180º − (30º + 70º) = 180º − 100º = 80º

Finding side ‘a’:

Finding side ‘c’:

12836.2°70sin°30sin4

sinsin

sinsin

=

=

=

=

BAba

Bb

Aa

b = 4

C

a

B = 70° c

A = 30°

3. Use the sine rule again to find the remaining unknown side.

2. Use the sine rule to find either of the two unknown sides.

1. Use the ‘sum of internal angles’ rule to find the unknown angle.

The triangle is now solved. This diagram shows all of the sides and angles:

C = 80° b = 4

a = 2.12836

B = 70° A = 30° c = 4.19204

1) Six triangles are shown below. In each case, two angles and one side are given. For each triangle, find the unknown angle and sides. Show all working steps. Use Maths Helper Plus to correct your work after solving each triangle.

To correct your work... 2) Start Maths Helper Plus and load the ‘R2 - Triangle_sover_AAS.mhp’ document. This document shows the working steps and a diagram for solving triangles when given two angles and one side.

b

85°

a

c

a = 7

b C = 24°

B

A = 25°

C

B = 28°

12° 40

c = 8 A = 62°

(c)

(b)

(a)

35°

6

73° (e)

(d)

40° 6

90° 55

144°

13°

(f)

3) Display the triangle solver options box Double click the mouse in the border to the left of the calculations. ( This area is shaded in the diagram below.) The triangle solver options box will display its 'Lengths & Angles' tab...

IMPORTANT Do not use the degree operator ° for angles, eg type 35 degrees as just 35, not as 35°.

Click the 'Clear' button to remove the previous triangle, then enter the two known angles and side length into the white edit boxes. Click the 'Apply' button at the bottom of the edit box. The calculated values will display on the options box. Click the 'OK' button to close the options box. The calculations and triangle diagram will be displayed on your screen. NOTE: If the diagram becomes too big for your computer screen, press the ‘F10’ key to make it smaller. To make the diagram bigger, hold down ‘Shift’ while you press ‘F10’.

Name: Class: p 51


Trigonometry - Activity 25

x

y(1,3)

(-2,-2)

(3,-1

General Triangle Solution: Given three (x,y) points. Plot three points on the (x,y) plane and join them with lines. If the points are not in the same straight line, you will have created a triangle. This diagram shows the triangle created by the three points (1,3), (-2,-2) and (3,-1): The distance formula can be used to find the distance between two (x,y) points. For example, consider the points (x1, y1) and (x2, y2). The distance formula gives the distance 'd' between them as follows:

212

212 )()( yyxxd −+−=

We can use the distance formula to find the distance between each pair of points making up our triangle. These distances are the lengths of the three sides. For the example above, the points are: (1,3), (-2,-2) and (3,-1), so we calculate the side lengths by taking these points two at a time: For the side joining (1,3) and (-2,-2) we have:

83095.534

)32()12(

)()(22

212

212

==

−−+−−=

−+−= yyxxd

Similarly, the side joining (1,3) and (3,-1) is 4.47214, and the side joining (-2,-2) and (3,-1)is 5.09902 We now know the three side lengths of the triangle, as shown below:

Once the three side measurements are known, then the internal angles 'A', 'B' and 'C' can be found using the procedure for a triangle when given the three side lengths. In summary, this method is as follows:

1. Use the cosine rule to find the largest angle (which is opposite the largest side). 2. Use the sine rule to find one of the remaining angles. 3. Use the ‘sum of internal angles’ rule to find the remaining angle.

For our example, the final result is as follows:

1) Six triangles are defined by the (x,y) points given below. Plot each set of points on the graph grid below, then find the unknown angle and sides. Show all working steps. Use Maths Helper Plus to correct your work after solving each triangle.

(a) (0,2) (2,5) (5,2) (b) (-2,-3) (-2,-5) (5,-3) (c) (1,5) (-5,4) (-2,2) (d) (1,-1) (-5,-3) (5,-2) (e) (-5,3) (-4,-2) (-1,1) (f) (5,0) (-1,2) (2,-1)

x

y

-5 -4 -3 -2 -1 0 1 2 3 4 5-5

-4

-3

-2

-1

0

1

2

3

4

5

To correct your work... 2) Start Maths Helper Plus and load the ‘R2 - Triangle_sover_3_points.mhp’ document. This document shows the working steps and a diagram for solving triangles when given three (x,y) points. 3) Display the triangle solver options box Double click the mouse in the border to the left of the calculations. ( This area is shaded in the diagram below.) The triangle solver options box will display its '3 Points' tab...

Click the 'Clear' button to remove the previous triangle, then enter the three (x,y) points into the white edit boxes. Click the 'Apply' button at the bottom of the edit box. The calculations will appear on the text view, and the diagram on the graph view of Maths Helper Plus. Click the 'OK' button to close the options box and view the calculations and diagram. NOTE: If the diagram becomes too big for your computer screen, press the ‘F10’ key to make it smaller. To make the diagram bigger, hold down ‘Shift’ while you press ‘F10’.

Documents

MHP Resource Set 2 - teachers choice software Set 2.pdf · Introduction The Maths Helper Plus Resource Set 2 consists of 25 classroom mathematics activities in trigonometry, along