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Geometry 8.7 Applications of Right Triangle Trigonometry.

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Presentation on theme: "Geometry 8.7 Applications of Right Triangle Trigonometry."— Presentation transcript:

1 Geometry 8.7 Applications of Right Triangle Trigonometry

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

3 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.

4 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 horizontal angle of elevation angle of depression

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

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

7 2.A ladder that is 20ft. 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?

8 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? ladder building 20 x 75˚ 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. cos

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

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

11 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 85m of string has been let out.

12 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 85 x 55˚ kite 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 sin

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

14 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 depression 38º 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 tan Alternate Interior Angles are congruent

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


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