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

10
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.

description

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. - PowerPoint PPT Presentation

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

Page 1: Use the 3 ratios –  sin ,  cos  and  tan  to solve application problems.

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.

Page 2: Use the 3 ratios –  sin ,  cos  and  tan  to solve application problems.

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

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

Angle of Elevation/Depression

Balloon

You

Angle of depression

Angle of elevation

Sometimes when we use right triangles to model real-life situations, we use the terms angle of elevation and angle of depression.

If you are standing on the ground and looking up at a hot air balloon, the angle that you look up from ground level is called the angle of elevation.

If someone is in the hot air balloon and looks down to the ground to see you, the angle that they have to lower their eyes, from looking straight ahead, is called the angle of depression.

Page 4: Use the 3 ratios –  sin ,  cos  and  tan  to solve application problems.

Angle of Elevation/Depression

If you look up 15º to see the balloon, then the person in the balloon has to look down 15º to see you on the ground.

Notice that in this situation, the one of the legs that forms the right angle is also the height of the balloon.

Angle of elevation = Angle of depression.

Balloon

You

Angle of depression = 15º

Angle of elevation= 15º

Page 5: Use the 3 ratios –  sin ,  cos  and  tan  to solve application problems.

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

Page 6: Use the 3 ratios –  sin ,  cos  and  tan  to solve application problems.

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 will Coach Jarvis have to crawl to get to the front door when he falls off the ladder (assuming he falls to the base of the ladder)?

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

Page 7: Use the 3 ratios –  sin ,  cos  and  tan  to solve application problems.

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

Page 8: Use the 3 ratios –  sin ,  cos  and  tan  to solve application problems.

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

Page 9: Use the 3 ratios –  sin ,  cos  and  tan  to solve application problems.

5. A 5.50 foot person standing 10 feet from a street light casts a 24 foot shadow. What is the height of the streetlight?

5.5

14 shadowx˚

tan x˚ = 5.5

14

x° ≈ 21.45°About 9 ft.

tan 21.4524

height

10

Page 10: Use the 3 ratios –  sin ,  cos  and  tan  to solve application problems.

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