Question

An observer, 1.6 m tall is 30 √3 m away from a tower the angle of elevation from his eye to the top of the tower is 30°. Determine the height of the tower.

Answer 1

the answer is 31.6mts

Answer 2

Answer:

Height of Tower is 31.6 m

Step-by-step explanation:

Given: Height of Observer, AB = 1.6 m

           Distance between tower and observer, CB = 30√3 m

           Angle of elevation = 30°

To find: Height of tower

from figure,

AE = CB = 30√3 m

In ΔADE,

using trigonometric ratio,

$tan\,30^{\circ}=\frac{DE}{AE}$

$\frac{1}{\sqrt{3}}=\frac{DE}{30\sqrt{3}}$   ( value of $tan\,30^{\circ}=\frac{1}{\sqrt{3}}$ )

$DE=\frac{30\sqrt{3}}{\sqrt{3}}$

DE = 30 m

Height of Tower = DE + CE = 30 + 1.6 = 31.6 m  (AB = CE)

Therefore, Height of Tower is 31.6 m