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4.2a: Right Triangle Trigonometry p. 412-419 GSE’s Covered M(G&M)–10–2 Makes and defends conjectures, constructs geometric arguments, uses geometric properties, or uses theorems to solve problems involving ang lines, polygons, circles, or right triangle ratios (sine, cosine, tangent) within mathematics or across disciplines or contexts (e.g., Pythagorean Theorem, Triangle Inequality Theorem). y: M(G&M)–10–9 Solves problems on and off the coordinate plane involving distance, midpoint, perpendicular and parallel lines, or slope.

4.2a: Right Triangle Trigonometry p. 412-419 GSE’s Covered Primary: M(G&M)–10–2 Makes and defends conjectures, constructs geometric arguments, uses geometric

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4.2a: Right Triangle Trigonometry

p. 412-419GSE’s Covered

Primary: M(G&M)–10–2 Makes and defends conjectures, constructs geometric arguments, uses geometric properties, or uses theorems to solve problems involving angles,lines, polygons, circles, or right triangle ratios (sine, cosine, tangent) within mathematics or across disciplines or contexts (e.g., Pythagorean Theorem, Triangle Inequality Theorem).

Secondary: M(G&M)–10–9 Solves problems on and off the coordinate plane involving distance, midpoint, perpendicular and parallel lines, or slope.

Using the reference angle for the right triangles above, identify: adjacent side, opposite side, hypotenuse.

Reference angle- an acute angle used in the right triangle

SOHCAHTOA

hyp

oppref

angle) sin(

All are sides of right triangles

hyp

aref

djangle) cos(

adj

oref

ppangle) tan(

Replace thisWith either the angleOr variable

What does it mean?

hyp

opprefangle )sin(

The sine of the reference angle is the ratio of the opposite side to the hypotenuse of a right triangle.

x

The angle we are talking about

The opposite side to the angle we are talking about

Always the hypotenuse in a right triangle

8 in9 in

So, sin x = 9

8

Lets solve this equation

x

A B

C

4 in

10 in

hyp

oppx

sin

10

4sin x

To solve for the angle, we need to get rid of sin

To get rid of sin and solve for the angle we use on both sides

1sin

10

4)(sinsin)(sin 11 x

10

4sin x

10

4sin 1x

24x Which means the angle is about 24 degrees

50

6 in x

Solve for x

Label the information you have in the triangle

Reference angle

AdjacentSide to The refangle

hypotenuse

If we have the Adjacent side and the Hypotenuse, think SOHCAHTOA

hyp

ax

djcos

x

650cos Now solve

For x

)( 6

50cos)( xx

x Multiple both sides by x

650cos)( x

50cos

6)( x Divide both sides by

Cos 50

inx 33.9Which means the hypotenuse is 9.3 in

70

8 ft

X ft

Solve for x

Label the information you have in the triangle

If we have the Opposite side and the Adjacent, think SOHCAHTOA

Adjacent side to the ref angle

Opposite side to the ref angle

adj

ox

pptan

8

70tan

x

)8(8

70tan)8(

x Multiply both sides by 8

x70tan)8(You have x alone, so evaluate 8 tan 70

ftx 22 So the opposite side is approximately 22 ft

Example on the coordinate plane

A (8,2)

B (4,5)

C (7,9) ABCin Cm Find

Secondary: M(G&M)–10–9 Solves problems on and off the coordinate plane involving distance, midpoint, perpendicular and parallel lines, or slope.

Primary: M(G&M)–10–2

Solve for the missing sides of the triangle using 2 different methods. Show all work

NECAP released Item 2007

Find the area of the triangle

Find the Volume of the Prism

Phil stands on the sidewalk of a road. Phil’s favorite pizza restaurant is on the other side of the road. His estimated line of sight to the pizza place is 43 degrees. He needs to go to the post office at some point which is 120 feet up the road he is standing on. The line of sight from the post office to the pizza place is 90 degrees.

How far of walk would it be for Phil from his original position to the pizza place?

How far is the walk from the post office to the pizza place?

Homework