An observer 1.6 m tall is 20√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The heights of the tower is:

Q: An observer 1.6 m tall is 20√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The heights of the tower is:

Given Observer's eye height = 1.6 m. Horizontal distance to tower = 20√3 m. Angle of elevation = 30°. Compute tower height H tan 30° = (H − 1.6) / (20√3) = 1/√3⇒ H − 1.6 = 20 ⇒ H = 21.6 m

Difficulty: Easy

20.0 m
21.6 m
23.2 m
24.0 m

Correct Answer: 21.6 m

Explanation:

Given

Observer's eye height = 1.6 m.
Horizontal distance to tower = 20√3 m.
Angle of elevation = 30°.

Compute tower height H
tan 30° = (H − 1.6) / (20√3) = 1/√3⇒ H − 1.6 = 20 ⇒ H = 21.6 m

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