Warm upFind the missing side.

Skills Check

CCGPS GeometryDay 4

UNIT QUESTION: What patterns can I find in right triangles?Standard: MCC9-12.G.SRT.6-8

Today’s Question:How do we use trig ratios to solve real world problems?Standard: MCC9-12.G.SRT.6-8

CCGPS Geometry

Applications of Right Triangle Trigonometry

Use the 3 ratios – sin, cos and tan to solve application problems.

Solving Word Problems

Choose the easiest ratio(s) to use based on what information you are given in the problem.

Draw a PictureWhen solving math problems, it can be very helpful to draw a picture of the situation if none is given.

Here is an example.

Find the missing sides and angles for Triangle FRY. Given that angle Y is the right angle, YR = 68, and FR = 88.

RY

F

68

88r

The picture helps to visualize what we know and what we want to find!

1. From a point 80m from the base of a tower, the angle of elevation is 28˚. How tall is the tower?

80

28˚

x

Using the 28˚ angle as a reference, we know opposite and adjacent sides.

Use tan 28˚ = x

80oppadj

80 (tan 28˚) = x

80 (.5317) = x x ≈ 42.5 About 43 m

tan

2. A ladder that is 20 ft is leaning against the side of a building. If the angle formed between the ladder and ground is 75˚, how far is the bottom of the ladder from the base of the building?

ladd

er

bu

ildin

g20

x

75˚

Using the 75˚ angle as a reference, we know hypotenuse and adjacent side.

Use cos 75˚ = x

20adjhyp

20 (cos 75˚) = x

20 (.2588) = x x ≈ 5.2 About 5 ft.

cos

3. When the sun is 62˚ above the horizon, a building casts a shadow 18m long. How tall is the building?

x

18shadow

62˚

Using the 62˚ angle as a reference, we know opposite and adjacent side.

Use tan 62˚ = x18

oppadj

18 (tan 62˚) = x

18 (1.8807) = xx ≈ 33.9 About 34 m

tan

4. A kite is flying at an angle of elevation of about 55˚. Ignoring the sag in the string, find the height of the kite if 85m of string have been let out.

string

85x

55˚

kite

Using the 55˚ angle as a reference, we know hypotenuse and opposite side.

Use sin 55˚ = x

85opphyp

85 (sin 55˚) = x

85 (.8192) = x x ≈ 69.6 About 70 m

sin

5. A 5.50 foot person standing 10 feet from a street light casts a 14 foot shadow. What is the height of the streetlight?

5.5

14 shadowx˚

tan x˚ = 5.5

14

x° ≈ 21.45°About 9.4 ft.

tan 21.4524

height

10

Depression and Elevation

If a person on the ground looks up to the top of a building, the angle formed between the line of sight and the horizontal is called the angle of elevation.

If a person standing on the top of a building looks down at a car on the ground, the angle formed between the line of sight and the horizontal is called the angle of depression.

horizontal

line of sight

horizontalangle of elevation

angle of depression

6. The angle of depression from the top of a tower to a boulder on the ground is 38º. If the tower is 25m high, how far from the base of the tower is the boulder?

25

x

angle of depression38º

38º

Using the 38˚ angle as a reference, we know opposite and adjacent side.

Use tan 38˚ = 25/x oppadj

(.7813) = 25/x

X = 25/.7813 x ≈ 32.0 About 32 m

tan

Alternate Interior Angles are congruent