Solutions of Oblique Triangles

7/28/2019 Solutions of Oblique Triangles

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TRIANGLES

Solutions of Oblique Triangles


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Conditions that Determine a Triangle

A triangle is said to be determined whenthe measures of three parts are given.

Three angles do not serve to determinea triangle. A triangle is determined inany of the following cases:


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Cases:Case 1. Law of Sines.

Given two angles and one side. Obtain the thirdangle, and then use the sine law.

Case 2. Ambiguous Case.

Given two sides and the angle opposite one of

them. Use the sine law. Examine the possibility ofno solution, one or two solutions.

Case 3. Given two sides and the included angle.

Use the cosine law for the third side and the sine

law for the angles. Find the smaller angle first toavoid the possibility of an obtuse angle.

Case 4. Given three sides,

Use the cosine law for each angle in turn.


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Oblique Triangles

c = 20

b a

A

C

B

0

300

65

1.


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2.

b = 24 a = 24

C = 18A B

C


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Oblique Triangles

An oblique triangle is a triangle which doesnot contain a right angle of 900. It contains

either three acute angles, or two acuteangles and one obtuse angle.

All acute angles One obtuse angle


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A triangle is uniquelydetermined when three parts,

not angles are known. Thusany triangle problem may fallunder any one of the

following cases:


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CasesCase I. Given one sideand two

angles

Case II. Given two sidesand an

angleopposite one of them

Case III. Given two sidesand theincluded angle.

Case IV. Given the three sides


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Law of Sines

In any triangle, the sides areproportional to the sines of the

opposite angles, that is,a b c

-------- = -------- = --------

sin sin sin


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Note:

We say,

sin a sin b sin c

------ =----; -------- =----; ------- =----

sin b sin c sin a


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Proof ( The given angle is acute )

Let ABC be any oblique triangle, In the figureangles and are acute. Let CDAB anddenote h = CD.

A B

C

D

hb

In the right triangle ACD above,h = b sin


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Proof ( The given angle is obtuse )

Let ABC be any oblique triangle,In the figure angleis obtuse. Let CDAB and denote h = CD.

A B

C

D

In the right triangle ACD above,h = a sin


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Thus,

h = h

b sin = a sin

sin a

---------- = ----------

sin b


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Similarly,

we obtain

a c b c

------- = -------- or --------- = --------

sin sin sin sin

a b c

---------- = ----------- = -------------

sin sin sin

Therefore,


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Cont.

Solutions ofOblique Triangles


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Case I

Given one side and two angles

Let a, , be given.

To find , use= 1800 ( + ).

To find b, use

sin b

------- = -------

sin a

or

a sin

b = --------

sin

To find c, use

sin c

------- = -------

sin a

or

a sin

c = -------

sin

a

b

c


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Example 1Solve ABC, given c = 20, = 300, = 650

c = 20300 650

A B

C

= 1800 - ( 300 + 650 ) = 850

ac sin

a = ------------

sin

20 sin 300

= ----------------

sin 850

= 10.4

c sin

b = ------------

sin

20 sin 650

= ------------

sin 850

= 18.20


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Example 2

A and B are two points on opposite bankof a river. From A a line AC = 270 ft. islaid off and the angles CAB = 1200 and

ACB = 450 are measured. Find thelength of AB.

A

B

A

C

b = 270 ft.1200

450

c = ?


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Solution

= 1800 ( 1200 + 450 ) = 150

b sin

c = ------------

sin

270 sin 450

= -------------

sin 15

737.65 ft.=


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Case II

Given two sides and the angle opposite one of

them. Let b, c and are given.

From

sin c

-------- = -----,

sin b

c sin

sin = ----------

b

If sin > 1 no angle is determinedIf sin = 1, = 900, a right triangle is determined

If sin < 1, two angles are determined:a.) an acute angle and

b.)an obtuse angle = 1800 -

Thus, there may be one or two triangles determined.


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Geometrically,

Let us discuss several special cases ofCase II, given two sides

angle opposite one of them.

a b ba

Let a, b, and be the given parts.


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ab

a

c

Given two sides and the angle opposite one of them.

Examples:


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The given angleis acute (1st)

In the figure: when b < AD and AD= c sin ,then the arc does not meet BE, hence notriangle is determined.

bc

A

DB E

c > b

h > b

h Thus, nois formed


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The given angleis acute (2nd)

In the figure, when b = AD, then the arcis tangent to BE, hence one right

triangle is determined.

B D

A

c b

E

c > b

c > h

b = h, h = c sin


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The given angleis acute (3rd)

In the figure, when b > AD but b < c, thenthe arc meets BE in two points C and C.Hence, two triangles ABCand ABC aredetermined. In ABC, is an acute anglewhile in ABC, = 1800 is an obtuseangle.

b

A

B C

c

DC

c > bc > b > h

hb


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The given angleis acute (4th)

In the figure, when b > AD and b = c, then

the arc meets BE in C and B. Hence oneisosceles triangle is determined.

bc = b

A

B C

D E

c = bb > h

h


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The given angleis acute (5th)

In the figure, when b > c, then the arcmeets BE in C and BE extended in C.Since ABC does not contain the given

angle , only one triangle is determined.

A

B C

bc

a E

C

b > c


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The given angle is obtuse(1st)

In the figure, when b < c or b = c, thenno triangle is determined.

A

B E

c

b b < c

b = c

Thus, nois formed


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The given angle is obtuse(2nd)

In the figure, when b > c, then only onetriangle is determined

A

B C

c

a

b

E

The arc intersects BE in two points but there is only

one which determines a triangles containing.


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1. When the angle is acute, there will be

a.) one solutionif the side opposite thegiven angle is greater than or equal to

the other given side, and

b.) no solution, one solution or

two solutionsi f the side opposi te the

given angle is less than the other given s ide.

c sin

sin = -------------

b

This sub case is determ ined by comput ing

To summarize this ambiguous case:


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2. When the angle is acute, there will be

b.) one solutionif the side opposite thegiven angle is greater than the other given

side.

a.) no solutionif the side opposite the givenangle is less than or equal to the other given side.


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Example 3

a.) When b = 20, c = 10, = 300

there is one solution since isacute and b > c.

Answer:

b = 20c = 10300


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b.)When b = 10, c = 20, = 300

Answer:

there is no solution, one solution or

two solution.


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c.)When b = 25, c = 15, = 1300

Draw the figure:

Answer:

there is one solution.

b = 25

c = 15 1300


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d.)When b = 25, c = 35, = 1300

Answer:

there is no solution


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Example 4

Solutions:

Since is acute and c > b,

there is only one solution.

For :

b sin

s in = --------------

c

80 sin 620

= ------------------

128

= 0.5518

s in = 0.5518 ,

=

330

30

620b = 80

c =128

Solve ABC, given c = 128, b = 80, = 620.


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For: = 1800( + )

= 84030

b sin 80 sin 840

30

For a:a = --------------- = -------------------- =144.28

sin sin 33030

Example 4 (solution)


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Example 5

Solve ABC, given a = 425, c = 310, = 120050

120050

A

B

C

a = 425c = 310

Example 5 (solution)


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Example 5 (solution)Solve ABC, given a = 425, c = 310, = 120050

Solution:

Since is obtuse and a > c, there is one solution.

c sin 310 sin 120050

For : sin = ----------- = -----------------------, =38046

a 425

For : = 1800 ( + ) = 20024

a sin 425 sin 20024

For b: b = -------------- = -------------------- = 172.53

sin sin 120050


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Law of CosinesIn any ABC, the square of any side isequal to the sum of the squares of theother two sides minus twice the product of

the sides and the cosine of their includedangle, that is,

a2= b2+c22bc cos ,

b

2

= a

2

+c

2

2ac cos c2= a2+ b22ab cos .

C


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Proof1:

In the right triangleACD in the figure

In the right BCD in the figure,h = a sin and DB = a cos .

A

C

BD

bh

b2 = h2 + AD2

a

AD = ABDBThen,

c

= c - a cos

b2= a2sin2+ ( c a cos )2


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So,

b2=a2sin2+ c22 ac cos +a2cos2

Therefore,

b2= a2+ c22 ac cos .

b2

=a2

sin2

+ a2

cos2

+ c2

2 ac cosb2=a2(sin2+ cos2)+ c22 ac cos

b2= a2sin2+ ( c a cos )2

C


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Proof2:

In the right triangleACD in the figure

A

C

B

b2 = h2 + AD2

D

bh

In the right BCD in the figure,

a

h = a sin = a sin (1800-)

= a sin

BD = a cos


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Therefore:

b2 = h2 + AD2A D

bh

= a2sin2 + (c a cos )2

= a2sin2 + c22ac cos +a2cos2

= a2(sin2 + cos2)+ c22ac cos

b2 = a2 + c2 2ac cos

Note:The other formulas are obtainedby cyclic changes of the letters.

B


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CASE III

Given two sides and the included angleLet a, b, and are given.

To find c: use c2= a2+ b22ab cos .

To find : use

use

a sin sin = -------------

c

To find :

b sin sin = -------------.

c

To check, use + + = 1800.

B


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Example 6.

Solve ABC, given a = 150, b = 230, = 32020.

For c: c2= a2+ b22ab cos

= 1502+ 23022 (150)(230) cos 32020

c

2

= 17098.39c = 130.76a sin 150 sin 32020

For : sin =-----------=-------------------; = 37050c 130.76

For : = 1800( + )= 1800( 37050 + 32020)= 109050

A

B

C32020

a = 150

b = 230


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Example 7Two forces 20.5 kgs and 25.5 kgs act on

a body. If their directions make anangle of 60020 with each other, findthe magnitude of their resultant and the

angle which it makes with the largerforce.

20.5

25.5

6002020.5

25.5

A B

CD

C25.5D


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Solution:

In parallelogram ABCD,

A + B = C + D = 1800

20.5

25.560

0

20

20.5

A B

B= 180060020 =119040

In ABC, b2= a2 + c2 2 a c cos b2= 20.52+ 25.522(20.5)(25.5) cos 119040

b =39.85kgs. (magnitude)a sin 20.5 sin 119040

sin = ------------- = ----------------------, =26033b 39.85

a

c


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CASE IV

Given three sides


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Let a, b and c be given. Apply the law of cosines foreach of the angles. To find the angles

a2+ b2c2

cos =---------------------2ab

b2+ c2a2

cos = -----------------,2bc

a2

+ c2

b2

cos =-------------------,2ac

To check, use + + = 1800.


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Example 8.

Solve ABC, given a=26.4, b = 34.5, c = 52.8.

a=26.4b = 34.5

c = 52.8


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Solution:

b2+ c2- a2 34.52+ 52.8226.42

For : cos = ----------------- = ------------------------- ; = 250452bc 2(34.5)(52.8)

a2+ c2b2 26.42+ 52.8234.52

For : cos = ---------------- = --------------------------; = 340362ac 2 (26.4) (52.8)

a2

+ b2

c2

26.42

+ 34.52

52.82

For : cos = ----------------- = -------------------------------; = 119038

2ab 2(26.4)(34.5)

a=26.4b = 34.5

c = 52.8

Check: + + = 1800

E l 9


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Example 9.

An isosceles triangle has sides thatmeasure 24 cm., 24 cm and 18 cm.Find the measure of each angle.

a = 24b = 24

c = 18


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Solution:242+ 182242

cos = --------------------------- , =670582(24)(18)

182+ 242242cos = -------------------------, 67058

2 (18)(24)

242+ 242 - 182cos = -------------------------, = 44003

2 (24)(24)

Check: + + = 1800

a=24b=24

c = 18


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Mabuhay!

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Solutions of Oblique Triangles