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CHAPTER 6:

The Trigonometric Functions

6.1The Trigonometric Functions of Acute Angles

6.2Applications of Right Triangles

6.3Trigonometric Functions of Any Angle 6.4Radians, Arc Length, and Angular Speed 6.5Circular functions: Graphs and Properties

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Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

6.2

Applications of Right Triangles

∙ Solve right triangles.

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Example

As a hot-air balloon began to rise, the ground crew drove 1.2 mi to an observation station. The initial observation from the station estimated the angle

between the ground and the line of sight to the balloon to be 30º. Approximately how high was the balloon at that point? (We are assuming that the wind velocity

was low and that the balloon rose vertically for the first few minutes.)

Solution:

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Slide 6.2 - 5

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Example

Solution continued:

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Example

A paint crew has purchased new 30-ft extension ladders. The manufacturer states that the safest

placement on a wall is to extend the ladder to 25 ft and to position the base 6.5 ft from the wall. What angle does the ladder make with the ground in this position?

Solution:

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Slide 6.2 - 7

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Example

Solution continued:

Thus when the ladder is in its safest position, it makes an angle of about 75º with the ground.

Use a calculator to find the acute angle whose cosine is 0.26:

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Solve a Right Triangle

To solve a right triangle means to find the lengths of

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Slide 6.2 - 9

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Example

In triangle ABC, find a, b, and B, where a and b represent the lengths of sides and B represents the

measure of angle B. Here we use standard lettering for naming the sides and angles of a right triangle: Side a is opposite angle A, side b is opposite angle B, where a and b are the legs, and side c, the hypotenuse, is opposite angle C, the right angle.

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Example

Solution:

B

b

106.2

C A

a

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Slide 6.2 - 11

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Example

• Find the missing side lengths and angles.

50° A + 50° = 90°, so A = 40°

y/10 = sin(50°)

10 x y = 10sin(50°) = 7.7

x = 10cos(50°) = 6.4

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Example using inverse trig

functions

• Find the angles A and B and side b given the following right triangle.

• Find angle A. Use an inverse trig function to find

A. For instance A = sin-1(6/10) = 36.9°.

• Then B = cos-1 (6/10) = 53.1°.

• To find b: b2 = 102 - 62 -> b2 = 64 -> b = 8.

B

6

10

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Slide 6.2 - 13

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Example

House framers can use trigonometric functions to

determine the lengths of rafters for a house. They first choose the pitch of the roof, or the ratio of the rise over the run. Then using a triangle with that ratio, they

calculate the length of the rafter needed for the house. Jose is constructing rafters for a roof with a 10/12 pitch on a house that is 42 ft wide. Find the length x of the rafter of the house to the nearest tenth of a foot.

Pitch: 10/12 Rise: 10

Run: 12

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Example

Solution:

First find the angle θ

that the rafter makes with the side wall.

θ ≈ 39.8º

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Slide 6.2 - 15

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Example

Solution continued:

The length of the rafter for this house is approximately 27.3 ft.

x

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Angle of Elevation

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Slide 6.2 - 17

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Angle of Depression

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Example

In Telluride, CO, there is a free gondola ride that provides a spectacular view of the town and the

surrounding mountains. The gondolas that begin in the town at an elevation of 8725 ft travel 5750 ft to Station St. Sophia, whose altitude is 10,550 ft. They then

continue 3913 ft to Mountain Village, whose elevation is 9500 ft.

a) What is the angle of elevation from the town to Station St. Sophia?

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Slide 6.2 - 19

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Example

Label a drawing with the given information.

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Example

a) Difference in elevation of St. Sophia to town is 10,550 ft – 8725 ft or 1825 ft. This is the side opposite the angle of elevation θ.

Solution

Town

Station St. Sophia

Angle of elevation 5750 ft

1825 ft

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Slide 6.2 - 21

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Example

Using a calculator, we find that

Solution continued

The angle of elevation from town to Station St. Sophia is approximately 18.5º.

b) When parallel lines are cut by a transversal,

alternate interior angles are equal. Thus the angle of depression, β, from Station St. Sophia to Mountain Village is equal to the angle of elevation from

Mountain Village to Station St. Sophia.

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Example

Difference in elevation of Station St. Sophia and the elevation of Mountain Village is 10,550 ft – 9500 ft, or 1050 ft.

Solution continued

The angle of depression from Station St. Sophia to Mountain Village is approximately 15.6º.

Mountain Village Station St. Sophia

Angle of elevation

3913 ft 1050 ft

β

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Slide 6.2 - 23

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Bearing: First-Type

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Example

A forest ranger at point

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Slide 6.2 - 25

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Example

Solution:

The forest ranger at point A is about 14.5 mi from the fire. Since d is the side opposite 62.62º, use the tangent

function ratio to find d.

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Example

In U.S. Cellular Field, the home of the Chicago White Sox baseball team, the first row of seats in the upper deck is farther away from home plate than the last row of seats in the original Comiskey Park. Although there is no obstructed view in U.S. Cellular Field, some of the fans still complain about the present distance from home plate to the upper deck of seats. From a seat in the last row of the upper deck directly behind the

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Slide 6.2 - 27

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Example

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Example

Solution:

We know that θ1 = 29.9º and θ2 = 24.2º. The distance form home plate to the pitcher’s mound is 60.5 ft. In the drawing, we d1 be the viewing distance to home plate, d2 the viewing distance to the pitcher’s mound, h

the elevation of the last row, and x the horizontal

distance form the batter to a point directly below the seat in the last row of the upper deck.

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Slide 6.2 - 29

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Example

Solution continued:

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Example

Solution continued:

The distance to home plate is about 250 ft, and the distance to the pitcher’s mound is about 304 ft.

References

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