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Copyright © 2009 Pearson Education, Inc. 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 6.6Graphs of Transformed Sine and Cosine Functions

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Copyright © 2009 Pearson Education, Inc. 6.2 Applications of Right Triangles Solve right triangles. Solve applied problems involving right triangles and trigonometric functions.

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Slide Copyright © 2009 Pearson Education, Inc. Solve a Right Triangle To solve a right triangle means to find the lengths of all sides and the measures of all angles.

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Slide Copyright © 2009 Pearson Education, Inc. 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. B b C A a 61.7º

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Slide Copyright © 2009 Pearson Education, Inc. Example Solution: B b C A a 61.7º

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Slide Copyright © 2009 Pearson Education, Inc. 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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Slide Copyright © 2009 Pearson Education, Inc. Example Solution: First find the angle that the rafter makes with the side wall. ≈ 39.8º Use the cosine function to determine the length x of the rafter.

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Slide Copyright © 2009 Pearson Education, Inc. Example Solution continued: The length of the rafter for this house is approximately 27.3 ft. x 21 ft 39.8º

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Slide Copyright © 2009 Pearson Education, Inc. Angle of Elevation The angle between the horizontal and a line of sight above the horizontal is called an angle of elevation.

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Slide Copyright © 2009 Pearson Education, Inc. Angle of Depression The angle between the horizontal and a line of sight below the horizontal is called an angle of depression.

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Slide Copyright © 2009 Pearson Education, Inc. 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? b)What is the angle of depression from Station St. Sophia to Mountain Village?

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Slide Copyright © 2009 Pearson Education, Inc. Example Label a drawing with the given information. Solution

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Slide Copyright © 2009 Pearson Education, Inc. 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 . Solution Town Station St. Sophia Angle of elevation 5750 ft 1825 ft

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Slide Copyright © 2009 Pearson Education, Inc. 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, , from Station St. Sophia to Mountain Village is equal to the angle of elevation from Mountain Village to Station St. Sophia. ≈ 18.5º

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Slide Copyright © 2009 Pearson Education, Inc. 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 Angle of depression

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Slide Copyright © 2009 Pearson Education, Inc. Bearing: First-Type One method of giving direction, or bearing, involves reference to a north-south line using an acute angle. For example, N55ºW means 55º west of north and S67ºE means 67º east of south.

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Slide Copyright © 2009 Pearson Education, Inc. Example A forest ranger at point A sights a fire directly south. A second ranger at point B, 7.5 mi east, sights the same fire at a bearing of S27º23 ´ W. How far from A is the fire?

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Slide Copyright © 2009 Pearson Education, Inc. 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. Find the complement of 27º23 ´.

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Slide Copyright © 2009 Pearson Education, Inc. 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 batter, the angle of depression to home plate is 29.9º, and the angle of depression to the pitcher’s mound is 24.2º. Find (a) the viewing distance to home plate and (b) the viewing distance to the pitcher’s mound.

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Slide Copyright © 2009 Pearson Education, Inc. Example Solution:

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Slide Copyright © 2009 Pearson Education, Inc. Example Solution: We know that 1 = 29.9º and 2 = 24.2º. The distance form home plate to the pitcher’s mound is 60.5 ft. In the drawing, we d 1 be the viewing distance to home plate, d 2 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. Begin by finding x.

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Slide Copyright © 2009 Pearson Education, Inc. Example Solution continued: Use the tangent function with 1 = 29.9º and 2 = 24.2º:

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Slide Copyright © 2009 Pearson Education, Inc. Example Solution continued: The distance to home plate is about 250 ft, and the distance to the pitcher’s mound is about 304 ft. Then find d 1 and d 2 using the cosine function:

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