4-1 Right Triangle Trigonometry - Mr. Upright at …...the adjacent side 1. By the Pythagorean...

Find the exact values of the six trigonometric functions of θ.

1.

SOLUTION:

The length of the side opposite θ is 8 , the length of the side adjacent to θ is 14, and the length of the hypotenuseis 18.

2.

SOLUTION:

The length of the side opposite θ is 2 , the length of the side adjacent to θ is 13, and the length of the hypotenuse is 15.

3.

SOLUTION:

The length of the side opposite θ is 9, the length of the side adjacent to θ is 4, and the length of the hypotenuse is

.

4.

SOLUTION:

The length of the side opposite θ is 12, the length of the side adjacent to θ is 35, and the length of the hypotenuse is 37.

5.

SOLUTION:

The length of the side opposite θ is , the length of the side adjacent to θ is 26, and the length of the hypotenuse is 29.

6.

SOLUTION:

The length of the side opposite θ is 30, the length of the side adjacent to θ is 5 , and the length of the hypotenuse is 35.

7.

SOLUTION:

The length of the side opposite θ is 6 and the length of the hypotenuse is 10. By the Pythagorean Theorem, the

length of the side adjacent to θ is = or 8.

8.

SOLUTION:

The length of the side opposite θ is 8 and the length of the side adjacent to θ is 32. By the Pythagorean Theorem, the length of the hypotenuse is

= or 8 .

Use the given trigonometric function value of the acute angle θ to find the exact values of the five remaining trigonometric function values of θ.

9. sin θ =

SOLUTION:

Draw a right triangle and label one acute angle θ. Because sin θ = = , label the opposite side 4 and the

hypotenuse 5.

By the Pythagorean Theorem, the length of the side adjacent to θ is or 3.

10. cos θ =

SOLUTION:

Draw a right triangle and label one acute angle θ. Because cos θ = = , label the adjacent side 6 and the

hypotenuse 7.

By the Pythagorean Theorem, the length of the side opposite θ is or .

11. tan θ = 3

SOLUTION:

Draw a right triangle and label one acute angle θ. Because tan θ = = 3 or , label the side opposite θ 3 and

the adjacent side 1.

By the Pythagorean Theorem, the length of the hypotenuse is or .

12. sec θ = 8

SOLUTION:

Draw a right triangle and label one acute angle θ. Because sec θ = = 8 or , label the hypotenuse 8 and

By the Pythagorean Theorem, the length of the side adjacent to θ is = or 3 .

13. cos θ =

SOLUTION:

hypotenuse 9.

By the Pythagorean Theorem, the length of the side opposite θ is or 2 .

14. tan θ =

SOLUTION:

Draw a right triangle and label one acute angle θ. Because tan θ = = , label the side opposite θ 5 and

adjacent side 4.

15. cot θ = 5

SOLUTION:

Draw a right triangle and label one acute angle θ. Because cot θ = = 5 or , label the side adjacent to θ as

5 and the opposite side 1.

16. csc θ = 6

SOLUTION:

Draw a right triangle and label one acute angle θ. Because csc θ = = 6 or , label the hypotenuse 6 and the

side opposite θ as 1.

By the Pythagorean Theorem, the length of the adjacent side is or .

17. sec θ =

SOLUTION:

Draw a right triangle and label one acute angle θ. Because sec θ = = , label the hypotenuse 9 and the side

opposite θ as 2.

By the Pythagorean Theorem, the length of the opposite side is or .

18. sin θ =

SOLUTION:

Draw a right triangle and label one acute angle θ. Because sin θ = = , label the side opposite θ as 8 and

the hypotenuse 13.

Find the value of x. Round to the nearest tenth.

19.

SOLUTION: An acute angle measure and the length of the hypotenuse are given, so the sine function can be used to find the

length of the side opposite θ.

20.

SOLUTION:

An acute angle measure and the length of the side opposite θ are given, so the tangent function can be used to find the length of the side adjacent to θ.

21.

SOLUTION: An acute angle measure and the length of the hypotenuse are given, so the cosine function can be used to find the

length of the side adjacent to θ.

22.

SOLUTION:

An acute angle measure and the length of the side adjacent to θ are given, so the cosine function can be used to find the length of the hypotenuse.

23.

SOLUTION:

An acute angle measure and the length of the side opposite θ are given, so the sine function can be used to find thelength of the hypotenuse.

24.

SOLUTION:

An acute angle measure and the length of the side adjacent to θ are given, so the tangent function can be used to find the length of the side opposite θ .

25.

SOLUTION:

An acute angle measure and the length of the side opposite θ are given, so the tangent function can be used to find the length of the side adjacent to θ .

26.

SOLUTION:

27. MOUNTAIN CLIMBING A team of climbers must determine the width of a ravine in order to set up equipment to cross it. If the climbers walk 25 feet along the ravine from their crossing point, and sight the crossing point on the far side of the ravine to be at a 35º angle, how wide is the ravine?

SOLUTION:

An acute angle measure and the adjacent side length are given, so the tangent function can be used to find the length of the opposite side.

Therefore, the ravine is about 17.5 feet wide.

28. SNOWBOARDING Brad built a snowboarding ramp with a height of 3.5 feet and an 18º incline. a. Draw a diagram to represent the situation. b. Determine the length of the ramp.

SOLUTION: a. Draw a diagram of a right triangle and label one acute angle 18º and the opposite side 3.5 feet.

b. Because an acute angle measure and opposite side length are given, the sine function can be used to find the length of the hypotenuse.

Therefore, the length of the ramp is about 11.3 feet.

29. DETOUR Traffic is detoured from Elwood Ave., left 0.8 mile on Maple St., and then right on Oak St., which intersects Elwood Ave. at a 32° angle. a. Draw a diagram to represent the situation. b. Determine the length of Elwood Ave. that is detoured.

SOLUTION: a. Draw a right triangle with acute angle 32º and opposite side length 0.8 mile. Label the side opposite the 32º angleMaple St., the hypotenuse Oak St., and the adjacent side Elwood Ave.

b. Because an acute angle and the length of opposite side are given, the tangent function can be used to find the adjacent side length.

Therefore, the length of Elwood Ave. that is detoured is about 1.3 miles.

30. PARACHUTING A paratrooper encounters stronger winds than anticipated while parachuting from 1350 feet, causing him to drift at an 8º angle. How far from the drop zone will the paratrooper land?

SOLUTION: Because an acute angle and the length of the side that is adjacent to the angle are given, the tangent function can be used to find the length of the opposite side.

So, the paratrooper will land about 190 feet away from the drop zone.

Find the measure of angle θ. Round to the nearest degree, if necessary.

31.

SOLUTION:

Because the lengths of the sides opposite and adjacent to θ are given, the tangent function can be used to find θ .

32.

SOLUTION:

Because the length of the hypotenuse and side adjacent to θ are given, the cosine function can be used to find θ .

33.

SOLUTION:

Because the length of the hypotenuse and side opposite θ are given, the sine function can be used to find θ .

34.

SOLUTION:

35.

SOLUTION:

36.

SOLUTION:

37.

SOLUTION:

38.

SOLUTION:

39. PARASAILING Kayla decided to try parasailing. She was strapped into a parachute towed by a boat. An 800-foot line connected her parachute to the boat, which was at a 32º angle of depression below her. How high above the water was Kayla?

SOLUTION: The angle of elevation from the boat to the parachute is equivalent to the angle of depression from the parachute tothe boat because the two angles are alternate interior angles, as shown below.

Because an acute angle and the hypotenuse are given, the sine function can be used to find x.

Therefore, Kayla was about 424 feet above the water.

40. OBSERVATION WHEEL The London Eye is a 135-meter-tall observation wheel. If a passenger at the top of the wheel sights the London Aquarium at a 58º angle of depression, what is the distance between the aquarium andthe London Eye?

SOLUTION: The angle of depression from the top of the wheel to the aquarium is 58º, which means that the angle of elevation from the aquarium to the top of the wheel is also 58º because they are alternate interior angles. Draw a diagram of a right triangle and label one acute angle 58º and the opposite side 135 m.

Because an acute angle and the side length opposite the angle are given, the tangent function can be used to find x.

Therefore, the distance between the aquarium and the London Eye is about 84 meters.

41. ROLLER COASTER On a roller coaster, 375 feet of track ascend at a 55º angle of elevation to the top before the first and highest drop. a. Draw a diagram to represent the situation. b. Determine the height of the roller coaster.

SOLUTION: a. Draw a diagram of a right triangle and label one acute angle 55º and the hypotenuse 375 feet.

b. Because an acute angle and the length of the hypotenuse are given, you can use the sine function to find the length of the opposite side.

Therefore, the height of the roller coaster is about 307 feet.

42. SKI LIFT A company is installing a new ski lift on a 225-meter-high mountain that will ascend at a 48º angle of elevation. a. Draw a diagram to represent the situation. b. Determine the length of cable the lift requires to extend from the base to the peak of the mountain.

SOLUTION: a. Draw a diagram of a right triangle and label one acute angle 48º and the opposite side 225 meters.

b. Because an acute angle and the length of the side opposite the angle are given, you can use the sine function to find the length of the hypotenuse.

So, the company will need about 303 meters of cable.

43. BASKETBALL Both Derek and Sam are 5 feet 10 inches tall. Derek looks at a 10-foot basketball goal with an angle of elevation of 29°, and Sam looks at the goal with an angle of elevation of 43°. If Sam is directly in front of Derek, how far apart are the boys standing?

SOLUTION: Draw a diagram to model the situation. The vertical distance from the boys' heads to the rim is 10(12) – [5(12) + 10] or 50 inches. Label the horizontal distance between Sam and Derek as x and the horizontal distance between Sam and the goal as y .

From the smaller right triangle, you can use the tangent function to find y .

From the larger right triangle, you can use the tangent function to find x.

Therefore, Derek and Sam are standing about 36.6 inches or 3.1 feet apart.

44. PARIS A tourist on the first observation level of the Eiffel Tower sights the Musée D’Orsay at a 1.4º angle of depression. A tourist on the third observation level, located 219 meters directly above the first, sights the Musée D’Orsay at a 6.8º angle of depression. a. Draw a diagram to represent the situation. b. Determine the distance between the Eiffel Tower and the Musée D’Orsay.

SOLUTION:

a.

b. Because the angles of depression from the 1st

and 3rd levels to the Musée D’Orsay are 1.4° and 6.8°,

respectively, the angles of elevation from the Musée D’Orsay to the 1st and 3rd levels are also 1.4° and 6.8°, respectively.

Use the tangent function to write an equation for the smaller right triangle in terms of y .

Next, use the tangent function to write an equation for the larger right triangle in terms of y .

Set the equations that you found for each triangle equal to one another and solve for x.

Therefore, the distance between the Eiffel Tower and the Musée D’Orsay is about 2310 meters.

45. LIGHTHOUSE Two ships are spotted from the top of a 156-foot lighthouse. The first ship is at a 27º angle of depression, and the second ship is directly behind the first at a 7º angle of depression. a. Draw a diagram to represent the situation. b. Determine the distance between the two ships.

SOLUTION: a.

b. From the smaller right triangle, you can use the tangent function to find y .

From the larger right triangle, you can use the tangent fuction to find x.

Therefore, the distance between the two ships is about 964 feet.

46. MOUNT RUSHMORE The faces of the presidents at Mount Rushmore are 60 feet tall. A visitor sees the top of George Washington’s head at a 48º angle of elevation and his chin at a 44.76º angle of elevation. Find the height of Mount Rushmore.

SOLUTION: From the smaller triangle, you can use the tangent function to find y .

From the larger triangle, you can use the tangent function to find y , too.

Next, set the two equations equal to one another to solve for x.

Therefore, Mount Rushmore is about 500 feet tall.

Solve each triangle. Round side measures to the nearest tenth and angle measures to the nearest degree.

47.

SOLUTION: Use trigonometric functions to find b and c.

Because the measures of two angles are given, B can be found by subtracting A from

Therefore, b 16.5, c 17.5.

48.

SOLUTION: Use trigonometric functions to find y and z .

Because the measures of two angles are given, X can be found by subtracting Z from

Therefore, y 37.1, z 32.5.

49.

SOLUTION: Use the Pythagorean Theorem to find r.

Use the tangent function to find P.

Because the measures of two angles are now known, you can find Q by subtracting P from

Therefore, P ≈ 43°, Q 47°, and r 34.0.

50.

SOLUTION: Use the Pythagorean Theorem to find d.

Use the cosine function to find D.

Because the measures of two angles are now known, you can find E by subtracting D from

Therefore, D ≈ 77°, E 13°, and d 29.2.

51.

SOLUTION: Use trigonometric functions to find j and k .

Because the measures of two angles are given, K can be found by subtracting J from

Therefore, j 18.0, k 6.2.

52.

SOLUTION: Use the Pythagorean Theorem to find y .

Use the sine function to find W.

Because the measures of two angles are now known, you can find Y by subtracting W from

Therefore, and y 3.9.

53.

SOLUTION: Use trigonometric functions to find f and h.

Because the measures of two angles are given, H can be found by subtracting F from

Therefore, f 19.6, h 17.1.

54.

SOLUTION: Use the Pythagorean Theorem to find t.

Use the tangent function to find R.

Because the measures of two angles are now known, you can find S by subtracting R from

Therefore, and t 8.1.

55. BASEBALL Michael’s seat at a game is 65 feet behind home plate. His line of vision is 10 feet above the field. a. Draw a diagram to represent the situation. b. What is the angle of depression to home plate?

SOLUTION: a.

b. Michael’s angle of depression to home plate is equivalent to the angle of elevation from home plate to where he is sitting.

Use the tangent function to find θ.

Therefore, the angle of depression to home plate is about

56. HIKING Jessica is standing 2 miles from the center of the base of Pikes Peak, and looking at the summit of the mountain, which is 1.4 miles from the base. a. Draw a diagram to represent the situation. b. With what angle of elevation is Jessica looking at the summit of the mountain?

SOLUTION: a.

b. Use the tangent function to find the angle of elevation.

Therefore, the angle of elevation is about

Find the exact value of each expression without using a calculator.57. sin 60°

SOLUTION: Draw a diagram of a 30º-60º-90º triangle.

The length of the side opposite the 60° angle is x and the length of the hypotenuse is 2x.

58. cot 30°

The length of the side adjacent to the 30º angle is x and the length of the opposite side is x.

59. sec 30°

The length of the hypotenuse is 2x and the length of the side adjacent to the 30º angle is x.

60. cos 45°

The side length is x and the hypotenuse is x.

61. tan 60°

The length of the side opposite the 60º is x and the length of the adjacent side is x.

62. csc 45°

The length of the hypotenuse is x and the side length is x.

Without using a calculator, find the measure of the acute angle θ that satisfies the given equation.63. tan θ = 1

SOLUTION:

Because tan θ = 1 and tan θ = , it follows that = 1. In the triangle below, the side length

opposite an acute angle is 1 and the adjacent side length is also 1. So, = 1.

Therefore, θ = 45°.

64. cos θ =

SOLUTION:

Because cos θ = and cos θ = , it follows that = . In the triangle below, the side

length adjacent to the 30º angle is and the hypotenuse is 2. So, = .

65. cot θ =

SOLUTION:

Because cot θ = and cot θ = , it follows that = . In the triangle below, the side

length that is adjacent to the 60º angle is 1 and the length of the opposite side is . So, = or .

66. sin θ =

SOLUTION:

Because sin θ = and sin θ = , it follows that = . In the triangle below, the side

length opposite an acute angle is 1 and the hypotenuse is . So, = or .

67. csc θ = 2

SOLUTION:

Because csc θ = 2 and csc θ = , it follows that = 2 or . In the triangle below, the

hypotenuse is 2 and the side length that is opposite the 30º angle is 1. So, = or 2.

68. sec θ = 2

SOLUTION:

Because sec θ = 2 and sec θ = , it follows that = 2 or . In the triangle below, the

hypotenuse is 2 and the side length that is adjacent to the 60º angle is 1. So, = or 2.

Without using a calculator, determine the value of x.

69.

SOLUTION: Because an acute angle, the hypotenuse, and the length of the side adjacent to the angle are given, the cosine

function can be used to find x. Using the properties of triangles, you can find that cos 30º = .

70.

SOLUTION:

Because the triangle is a triangle, the legs are the same length.

71. SCUBA DIVING A scuba diver located 20 feet below the surface of the water spots a shipwreck at a 70º angle of depression. After descending to a point 45 feet above the ocean floor, the diver sees the shipwreck at a 57º angle of depression. Draw a diagram to represent the situation, and determine the depth of the shipwreck.

SOLUTION: First, draw a diagram to represent the situation.

Label the horizontal distance from the shipwreck to the point on the ocean floor below the diver as x. Label the vertical distance from the point 20 feet below sea level to the point 45 feet below sea level as y . Find the complementary angle for each angle of depression. Use the tangent function to write an equation for the smaller right triangle in terms of x.

Use the tangent function to write an equation for the larger right triangle in terms of x.

Set the two equations equal to one another to solve for y .

Therefore, the depth of the shipwreck is 20 + 35 + 45 or about 100 feet.

Find the value of cos θ if θ is the measure of the smallest angle in each type of right triangle.72. 3-4-5

SOLUTION: Draw a diagram of a 3-4-5 triangle. The smallest angle will be the angle opposite the side with a length of 3.

73. 5-12-13

74. SOLAR POWER Find the total area of the panel shown below.

SOLUTION:

The area of the panel is given by , where the width is 10 feet and the length is unknown. Notice that the length of the panel is also the hypotenuse of a right triangle with an acute angle of 55º and an opposite side length of 3.5 feet. The sine function can be used to find the hypotenuse of the triangle.

So, the length of the panel is 4.27 feet. Find the area.

Therefore, the area of the panel is 42.7 square feet.

Without using a calculator, insert the appropriate symbol >, cot 60º.

76. tan 60° cot 30°

SOLUTION:

Use a graphing calculator to find tan 60º and cot 30º. To find cot 30º, find .

tan 60º ≈ 1.732 cot 30º ≈ 1.732 Therefore, tan 60° = cot 30°.

77. cos 30° csc 45°

SOLUTION:

Use a graphing calculator to find cos 30º and csc 45º. To find csc 45º, find .

cos 30º ≈ 0.866 csc 45º ≈ 1.414 Therefore, cos 30° csc 60°.

80. tan 45° sec 30°

SOLUTION: tan 45° ○ sec 30°

Use a graphing calculator to find tan 45º and sec 30º. To find sec 30º, find .

tan 45º = 1 sec 30º ≈ 1.155 Therefore, tan 45°

1.

SOLUTION:

2.

SOLUTION:

3.

SOLUTION:

.

4.

SOLUTION:

5.

SOLUTION:

6.

SOLUTION:

7.

SOLUTION:

8.

SOLUTION:

= or 8 .

9. sin θ =

SOLUTION:

hypotenuse 5.

10. cos θ =

SOLUTION:

hypotenuse 7.

11. tan θ = 3

SOLUTION:

12. sec θ = 8

SOLUTION:

13. cos θ =

SOLUTION:

hypotenuse 9.

14. tan θ =

SOLUTION:

adjacent side 4.

15. cot θ = 5

SOLUTION:

16. csc θ = 6

SOLUTION:

17. sec θ =

SOLUTION:

opposite θ as 2.

18. sin θ =

SOLUTION:

the hypotenuse 13.

19.

20.

SOLUTION:

21.

22.

SOLUTION:

23.

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24.

SOLUTION:

25.

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26.

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31.

SOLUTION:

32.

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33.

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34.

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35.

SOLUTION:

36.

SOLUTION:

37.

SOLUTION:

38.

SOLUTION:

a.

SOLUTION: a.

47.

48.

49.

50.

51.

52.

53.

54.

SOLUTION: a.

58. cot 30°

59. sec 30°

60. cos 45°

61. tan 60°

62. csc 45°

SOLUTION:

64. cos θ =

SOLUTION:

65. cot θ =

SOLUTION:

66. sin θ =

SOLUTION:

67. csc θ = 2

SOLUTION:

68. sec θ = 2

SOLUTION:

69.

70.

SOLUTION:

73. 5-12-13

SOLUTION:

76. tan 60° cot 30°

SOLUTION:

77. cos 30° csc 45°

SOLUTION:

80. tan 45° sec 30°

1.

SOLUTION:

2.

SOLUTION:

3.

SOLUTION:

.

4.

SOLUTION:

5.

SOLUTION:

6.

SOLUTION:

7.

SOLUTION:

8.

SOLUTION:

= or 8 .

9. sin θ =

SOLUTION:

hypotenuse 5.

10. cos θ =

SOLUTION:

hypotenuse 7.

11. tan θ = 3

SOLUTION:

12. sec θ = 8

SOLUTION:

13. cos θ =

SOLUTION:

hypotenuse 9.

14. tan θ =

SOLUTION:

adjacent side 4.

15. cot θ = 5

SOLUTION:

16. csc θ = 6

SOLUTION:

17. sec θ =

SOLUTION:

opposite θ as 2.

18. sin θ =

SOLUTION:

the hypotenuse 13.

19.

20.

SOLUTION:

21.

22.

SOLUTION:

23.

SOLUTION:

24.

SOLUTION:

25.

SOLUTION:

26.

SOLUTION:

31.

SOLUTION:

32.

SOLUTION:

33.

SOLUTION:

34.

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35.

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36.

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37.

SOLUTION:

38.

SOLUTION:

a.

SOLUTION: a.

47.

48.

49.

50.

51.

52.

53.

54.

SOLUTION: a.

58. cot 30°

59. sec 30°

60. cos 45°

61. tan 60°

62. csc 45°

SOLUTION:

64. cos θ =

SOLUTION:

65. cot θ =

SOLUTION:

66. sin θ =

SOLUTION:

67. csc θ = 2

SOLUTION:

68. sec θ = 2

SOLUTION:

69.

70.

SOLUTION:

73. 5-12-13

SOLUTION:

76. tan 60° cot 30°

SOLUTION:

77. cos 30° csc 45°

SOLUTION:

80. tan 45° sec 30°

1.

SOLUTION:

2.

SOLUTION:

3.

SOLUTION:

.

4.

SOLUTION:

5.

SOLUTION:

6.

SOLUTION:

7.

SOLUTION:

8.

SOLUTION:

= or 8 .

9. sin θ =

SOLUTION:

hypotenuse 5.

10. cos θ =

SOLUTION:

hypotenuse 7.

11. tan θ = 3

SOLUTION:

12. sec θ = 8

SOLUTION:

13. cos θ =

SOLUTION:

hypotenuse 9.

14. tan θ =

SOLUTION:

adjacent side 4.

15. cot θ = 5

SOLUTION:

16. csc θ = 6

SOLUTION:

17. sec θ =

SOLUTION:

opposite θ as 2.

18. sin θ =

SOLUTION:

the hypotenuse 13.

19.

20.

SOLUTION:

21.

22.

SOLUTION:

23.

SOLUTION:

24.

SOLUTION:

25.

SOLUTION:

26.

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31.

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32.

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33.

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34.

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35.

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36.

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37.

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38.

SOLUTION:

a.

SOLUTION: a.

47.

48.

49.

Because the measures of two angles are now known, you c

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4-1 Right Triangle Trigonometry - Mr. Upright at …...the adjacent side 1. By the Pythagorean...

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