Trigonometry Real-World Applications 2

Question 1

A ship's navigator uses radar to detect an obstacle at a bearing of N 35° E and a distance of 12 nautical miles. How far north of the ship is the obstacle?

Accepted Answer

9.83 nautical miles

Answer

The northward distance is 12 × cos(35°) ≈ 12 × 0.819 ≈ 9.83 nautical miles.

Question 2

A Ferris wheel has a radius of 25 feet and its center is 30 feet above the ground. A rider starts at the bottom. After rotating 120°, what is the rider's height above the ground?

Accepted Answer

42.5 feet

Answer

Height = 30 + 25 × sin(120° − 90°) = 30 + 25 × sin(30°) = 30 + 12.5 = 42.5 feet.

Question 3

A contractor needs to build a ramp with a 7° incline to cover a horizontal distance of 20 feet. What is the length of the ramp surface?

Accepted Answer

20.15 feet

Answer

Ramp length = 20 / cos(7°) ≈ 20 / 0.9926 ≈ 20.15 feet.

Question 4

Two forces of 80 N and 60 N act on an object at an angle of 90° to each other. What is the magnitude of the resultant force?

Accepted Answer

100 N

Answer

For perpendicular forces, the resultant = √(80² + 60²) = √(6400 + 3600) = √10000 = 100 N.

Question 5

A surveyor stands at point A and measures an angle of elevation of 28° to the top of a hill. Standing 150 feet closer at point B, the angle of elevation is 42°. What is the height of the hill (to the nearest foot)?

Accepted Answer

183 feet

Answer

Using two angle-of-elevation equations: h = 150 × tan(28°) × tan(42°) / (tan(42°) − tan(28°)) ≈ 183 feet.

Trigonometry Practice Test

Trigonometry Practice Test

Trigonometry Real-World Applications 2