developed by Vicki Borlaug Walters State Community College Summer 2008

1. Word Problems: Finding a Side of a Right Triangle (given a side and an angle) developed by Vicki Borlaug Walters State Community College Summer 2008 Note to Instructor: These word problems do not require Law of Sines or Law of Cosines.

2. Question developed by V. Borlaug WSCC, 2008 Not drawn to scale. Example 1.) A straight boat ramp is being designed to cover a horizontal distance of 150 feet with an angle of elevation of 3º. Find the length of the boat ramp.

3. Example 2.) A hot air balloon rises vertically. Katie is standing on the level ground 20 feet from a point on the ground where the balloon was launched. She points her camera at the balloon and takes a picture. When Katie takes the picture the camera is 5.5 feet off the ground and has a 80º angle of elevation . Find the height of the hot air balloon when the picture is taken. Question developed by V. Borlaug WSCC, 2008 Not drawn to scale.

4. Example 3.) A crane has a 100 foot arm with a hook at the end of the arm. This crane is designed so that the angle of elevation of the arm can be changed. When the crane’s hook is attached to an object on the ground, the arm’s angle of elevation is 5º. The crane’s arm then rotates upward and raises the object to a position that gives the arm a 25º angle of elevation. Find the height the crane has lifted the object. HINT: Find the tip of the arm’s initial distance from horizontal. Then find the distance from horizontal after the object has been lifted. Use these to find the height the object has been lifted. Not drawn to scale. Question developed by V. Borlaug WSCC, 2008

5. Example 4.) Derek is scuba diving in still water. Starting from the surface he dives in a straight line with a 65º angle of depression. Derek is traveling at a constant rate of 25 feet per minute along this straight line path. a.) Find the distance Derek has traveled along this straight line path three minutes into the dive. b.) Find Derek’s vertical depth three minutes into the dive. c.) Find the rate at which Derek’s vertical depth is changing as he descends. Question developed by V. Borlaug WSCC, 2008

6. Example 5.) A river runs between a tree and a lamp post. The tree is directly north of the lamp post. The surveyor is 324 feet east of the lamp post. He measures an angle of 11.3º from the tree to the lamp post. Find the distance from the surveyor to the tree. Question developed by V. Borlaug WSCC, 2008 Not drawn to scale.

7. Example 6.) Jessica is standing 75 feet from the base of a vertical tree. She is 5 ½ feet tall and her eyes are 4 inches from the top of her head. It takes a 38º angle of elevation for Jessica to look at the top of the tree. (Use four decimal place accuracy.) a.) Find the height of the tree in feet. b.) Find the height of the tree in meters. NOTE: One meter is the length equal to 1,650,763.73 wavelengths in a vacuum of the orange-red radiation of krypton 86. One meter also equals 39.37 inches. Ref: The American Heritage Dictionary of the English Language, American Publishing Co., Inc., 1969 Question developed by V. Borlaug WSCC, 2008 Not drawn to scale.

8. Example 7.) A helicopter is hovering 560 feet above a straight and level road that runs east and west. On the road east of the helicopter are a car and a truck. From the helicopter the angle of depression to the car is 48º and the angle of depression to the truck is 38º. Find the distance between the car and the truck. HINT: First find the distance from a point directly below the helicopter to the car. Then do the same for the truck. Use these to find the distance between the car and the truck. Question developed by V. Borlaug WSCC, 2008 Not drawn to scale.

9. Not drawn to scale. Question developed by V. Borlaug WSCC, 2008 Example 8.) A boat is due south of a light house. From his charts the captain knows that the light house is 0.76 miles east of the pier. The captain takes measurements from his boat and finds an angle of 55º from the light house to the pier. a.) Find the distance from the boat to the pier. b.) Find the distance from the boat to the light house.

10. Example 9.) A straight 67 inch ramp is being designed for a skate board course. a.) Find the vertical height required to give the skate board ramp a 75º angle of depression. b.) Find the vertical height required to give the skate board ramp a 65º angle of depression. c.) Find the vertical height required to give the skate board ramp a 55º angle of depression. Question developed by V. Borlaug WSCC, 2008

11. Example 10.) A helicopter takes off from ground level and goes 853 feet with an angle of elevation of 23º. The helicopter then changes course and goes 719 feet with and angle of elevation of 49º and then it hovers in this position. Find the height of the hovering helicopter.

12. (not dawn to scale) Example 11: A short building is 200 feet away from a taller building. Jessie is on the roof of the short building. To see the top of the taller building requires Jessie to look up with a 38 angle of elevation. To see the bottom of the taller building requires Jessie to look down with 12 angle of depression. Find the height of the taller building. (You may assume that Jessie’s height is negligible.) HINT: To solve begin by making two separate right triangles.

13. Example 12: Solar panels are being installed on a roof. The roof has an angle of elevation of 9.2º. Each solar panel is 7.20 feet long and 4.80 feet wide. The longer edge of each panel will be horizontally installed on the roof. The shorter edge of the panel will be installed with an angle of elevation 29.6º. a.) Find the vertical height of each solar panel relative to horizontal. b.) Find the vertical height of each solar panel relative to the roof directly the solar panel’s highest edge. Hint for part “b”: Using the two angles of elevation there here are two right triangles. First solve the larger right triangle completely.

14. Answers • 1) Cos 3 = 150 • X • X = 150.2 X 3° 150 2) Tan 80 = X 20 X = 113.4 113.4 + 5.5 = 118.9 X 80° 20

15. Answers 3) Sin 5 = X 100 X = 8.7 Sin 25 = Y 100 Y = 42.3 Height = 42.3 – 8.7 Height = 33.6 Feet 100 X 5° 100 Y 25°

16. Answers 65° X 75 c) 68.0 / 3 = 22.7 ft per min 4) a) 3 x 25 = 75 b) Sin 65 = X X = 68.0 75

17. Answers T X 11.3° L S 324 5) Cos 11.3 = 324 X X = 330.4

18. Answers X 38° 75 5 ft 2 inches 6) Height = 95.1764 + 5.1667 = 100.3431 feet or 100.3431 / 39.37 = 2.5488 meters Cos 38 = 75 X X = 95.1764 5’ 2” = 5.1667ft

19. Answers 48° H 38° 560 48 38 T C X Y 7) Tan 38 = 560 Y Y = 716.8 Tan 48 = 560 X X = 504.2 Distance = 716.8 – 504.2 = 212.6

20. Answers Sin 55 = .76 X X = .93 8) .76 P L Y 55° X B Tan 55 = .76 Y Y = .53

21. Answers Sin 65 = X 67 X = 60.7 X 67 75° 9) Sin 75 = X 67 X = 64.7 Sin 55 = X 67 X = 54.9

22. Answers 719 Y 49° 853 X 23° Sin 49 = Y 719 Y = 542.6 Sin 23 = X 853 X = 333.3 10) 333.3 + 542.6 = 875.9

23. Answers 200 X 38° 12° Y Tan 12 = Y 200 Y = 42.5 Tan 38 = X 200 X = 156.3 11) 156.3 + 42.5 = 198.8

24. Answers 4.80 X cos 29.6 = Y 4.8 Y = 4.17 Z 29.6° 9.2° Y 12) Sin 29.6 = X 4.8 X = 2.4 tan 9.2 = Z 4.17 Y = 0.7 2.4 – 0.7 = 1.7