Chapter 4 Pre-Calculus OHHS

4.8 Solving Problems with Trigonometry How to use right triangle trigonometry to solve well-known types of problems. 4-8

Vocabulary 4-8

Example 1 The angle of depression of a buoy from the top of the Barnegat Bay lighthouse 130 feet above the surface of the water is 6 . Find the distance x from the base of the lighthouse to the buoy. What do we do first? Draw a picture. 4-8

Example 1 4-8

Example 1 x 130 66 4-8

Now You Try P. 431, #3 4-8

Challenging Problem An observer in the top of a lighthouse determines that the angles of depression to two sailboats directly in line with the lighthouse are 3.5º and 5.75º. If the observer is 125 ft above sea level, find the distance between the boats. What’s our first step? Right! Draw a picture. 4-8

Draw A Picture 4-8

Label the Picture 125 ft 3.5º5.75º Most pictures are not drawn to scale. Alternate Interior Angles say that Angles of Depression and Angles of Elevation are congruent. x 4-8

Separate Into 2 Problems a 125 ft a 3.5º 125 ft b x 5.75º 3.5º 125 ft b 5.75º 4-8

Solve for a and b 125 ft a 3.5º 125 ft b 5.75º 4-8

Finish the Problem a= b= a 125 ft b x 5.75º 3.5º We are supposed to find x. x = a – b = – x = ft 4-8

Now You Try P. 432, #7 4-8

Example 3 A large, helium-filled penguin is moored at the beginning of a parade route awaiting the start of the parade. Two cables attached to the underside of the penguin make angles of 48  and 40  with the ground and are in the same plane as a perpendicular line from the penguin to the ground. If the cables are attached to the ground 10 feet from each other, how high above the ground is the penguin? 4-8

Example 3 h x 40  48  4-8

Example 3 4-8

Now You Try P. 432, #15 4-8

Example 4 A U.S. Coast Guard patrol boat leaves Port Cleveland and averages 35 knots (nautical mph) traveling for 2 hours on a course of 53  and then 3 hours on a course of 143 . What is the boat’s bearing and distance from Port Cleveland? Bearing is the angle measured clock- wise from due north. 4-8

Example 4 Find the distances. AB = 35∙2 = 70 BC = 35∙2 =

Now You Try P. 433, #17 4-8

Home Work P. 431 – 435, #2, 8, 12, 16, 18, 26, 37-42, 45,