Arithmetic: Heights and Distances — Solved MCQs Solved MCQs

Correct Answer: 30°

tan θ = height/shadow = 1/√3, so θ = 30°.

Correct Answer: 173 m

Distance = 100/tan 30° = 100√3 = 173.2 m, so 173 m.

Correct Answer: 21.6 m

Height above eye level = 20√3 x tan 30° = 20 m; adding the observer's 1.6 m gives 21.6 m.

Correct Answer: Data inadequate

Neither the tower's height nor the distance the man walked is given, so the distance from the base to P cannot be determined — data inadequate.

Correct Answer: 27.3 m

Height x (cot 30° - cot 45°) = 20 m, so height = 20/(√3 - 1) = 27.3 m.

Correct Answer: 9.2 m

cos 60° = 4.6/length, so length = 4.6/0.5 = 9.2 m.

Correct Answer: 273 m

The ships are 100/tan 30° = 173.2 m and 100/tan 45° = 100 m from the lighthouse, on opposite sides, so 173.2 + 100 = 273.2 m, i.e. 273 m.

Correct Answer: 32 kmph

At 45° the tower height equals the 60 m distance; at 30° the distance is 60/tan 30° = 103.9 m, so the boat covers 43.9 m in 5 seconds, about 31.6 kmph, i.e. approximately 32 kmph.

Correct Answer: 63.5 m

The objects are 150 m and 150/tan 60° = 86.6 m from the tower's foot, so they are 150 - 86.6 = 63.4 m apart, i.e. 63.5 m.

Correct Answer: 16 minutes 23 seconds

The car covers height x (√3 - 1) in 12 minutes, so the remaining height takes 12/(√3 - 1) = 16.39 minutes, i.e. 16 minutes 23 seconds.

Correct Answer: 10 meters

Read as depression angles from the tower top: the tower-pole distance is 15/tan 60° = 8.66 m, and the pole height is 15 - 8.66 x tan 30° = 10 m.