Geometry 8 7 Applications of Right Triangle Trigonometry

Geometry 8. 7 Applications of Right Triangle Trigonometry

Vocab • Add hypotenuse, the three proportions, Pythagorean Theorem and Converse, multiply fractions vs. proportions, 45 -45 -90 and 30 -60 -90 formulas, pythagorean triples, sine, cosine, tangent, angle of elevation, angle of depression to vocab list

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.

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. angle of depression of e lin h sig 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 t angle of elevation horizontal

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

1. From a point 80 m 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 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. If the angle formed between the ladder and the ground is 75 o, how far is the bottom of the ladder from the base of the 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? de r lad building 20 75˚ x Using the 75˚ angle as a reference, we know hypotenuse and adjacent side. Use cos 75˚ = 20 (cos 75˚) = x 20 (. 2588) = x x ≈ 5. 2 About 5 ft.

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

3. When the sun is 62˚ above the horizon, a building casts a shadow 18 m long. How tall is the building? x 18 62˚ shadow Using the 62˚ angle as a reference, we know opposite and adjacent side. Use 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 o. Ignoring the sag in the string, find the height of the kite if 85 m of string has been let out.

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 85 m of string have been let out. kite x st rin g 85 55˚ Using the 55˚ angle as a reference, we know hypotenuse and opposite side. Use sin 55˚ = 85 (sin 55˚) = x 85 (. 8192) = x x ≈ 69. 6 About 70 m

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

6. The angle of depression from the top of a tower to a boulder on the ground is 38º. If the tower is 25 m 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 38˚ = 25/x (. 7813) = 25/x X = 25/. 7813 x ≈ 32. 0 About 32 m

Homework • Finish Notes problems #5 and #7 on binder paper. Answers: #5: 83 m #7: 27 m Do pg. 318 #1 -6 • Vocab Test Tomorrow, Test Thursday •