An observer 2 m tall is 150 √3 m away from a tower. The angle of elevation from his eye to the top of the tower is 60°. The height of the tower is:

Correct option is A

Given

Height of observer = 2 m

Distance between observer and tower = 150√3 m

Angle of elevation from the eye of an observer = 60°

​Solution:  

Let the DC  is the height of the observer which is 2 m. AB is the height of the tower. ∠D is the angle of elevation from the eye of the observer towards the top of the tower. BC is the distance between the tower and observer which is 150√3.

In triangle AED

​$$tan 60° = \frac{AE}{DE}\\\ \\ √3 = \frac{x}{(150√ 3)}\\\ \\ √3 × 150√3 = x\\\ \\ 150 × 3 = x\\\ \\450 = x$$​​

AB = AE + EB

 AB = 450 + 2

 AB = 452m