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CCGPS Geometry Day 18 (8-30-13) 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?

Right Triangle Trigonometry CCGPS Geometry Applications of Right Triangle Trigonometry

Solving Word Problems Use the 3 ratios – sin, cos and tan to solve application problems. Choose the easiest ratio(s) to use based on what information you are given in the problem.

Draw a Picture When 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. 68 88 r 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? x 28˚ 80 Using the 28˚ angle as a reference, we know opposite and adjacent sides. Use tan tan 28˚ = 80 (tan 28˚) = x 80 (.5317) = x x ≈ 42.5 About 43 m

2. A ladder that is 20 ft is leaning against the side of a building 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? 20 building ladder 75˚ x Using the 75˚ angle as a reference, we know hypotenuse and adjacent side. Use cos cos 75˚ = 20 (cos 75˚) = x 20 (.2588) = x x ≈ 5.2 About 5 ft.

3. When the sun is 62˚ above the horizon, a building casts a shadow 18m long. How tall is the building? x 62˚ 18 shadow Using the 62˚ angle as a reference, we know opposite and adjacent side. Use tan tan 62˚ = 18 (tan 62˚) = x 18 (1.8807) = x x ≈ 33.9 About 34 m

4. A kite is flying at an angle of elevation of about 55˚ 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. kite 85 x string 55˚ Using the 55˚ angle as a reference, we know hypotenuse and opposite side. Use sin sin 55˚ = 85 (sin 55˚) = x 85 (.8192) = x x ≈ 69.6 About 70 m

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 x˚ 10 14 shadow tan x˚ = x° ≈ 21.45° About 9.4 ft.

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 angle of depression line of sight angle of elevation horizontal

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? 38º angle of depression 25 Alternate Interior Angles are congruent 38º x Using the 38˚ angle as a reference, we know opposite and adjacent side. Use tan tan 38˚ = 25/x (.7813) = 25/x X = 25/.7813 x ≈ 32.0 About 32 m

Classwork/Homework 6 Problem Application front WS Test Review Sheet