An observer 1.6 m tall is 20√3 m away from a tower. The angle of ele

1.	An observer 1.6 m tall is 20√3 m away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The height of the tower is: 1. 21.6 m2. 23.2 m3. 24.72 m4. None of these
Answer» Correct Answer - Option 1 : 21.6 m Given : Observer height is 1.6 m Height of tower = 20√3 Angle of elevation = 30° RQ = 20√ 3 Calculation : In ∆ PQR Tanθ = P/B Tan30° = PQ/RQ \(\frac{1}{{√ 3 }} = \frac{{PQ}}{{RQ}}\) PQ = 20 m Hence total height of tower = 20 + 1.6 = 21.6 m

An observer 1.6 m tall is 20√3 m away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The height of the tower is: 1. 21.6 m2. 23.2 m3. 24.72 m4. None of these

Answer» Correct Answer - Option 1 : 21.6 m

Given :

Observer height is 1.6 m

Height of tower = 20√3

Angle of elevation = 30°

RQ = 20√ 3

Calculation :

In ∆ PQR

Tanθ = P/B

Tan30° = PQ/RQ

\(\frac{1}{{√ 3 }} = \frac{{PQ}}{{RQ}}\)

PQ = 20 m

Hence total height of tower = 20 + 1.6 = 21.6 m