An observer at the top of a tower observes that two cars are running towards the foot of the tower at a distance of 120 m from each other making angles of depression α and β such that α > β and tan α = $$\sqrt3$$​ and tan β = $$\frac{1}{\sqrt3}$$​. Find the height of the tower.

Correct option is D

Given:

Two cars are 120 m apart and approaching the foot of a tower.

Observer at the top sees them at angles of depression α and β, with

$$\tan\alpha = \sqrt{3}, \ \ \tan\beta = \frac{1}{\sqrt{3}}$$​ and α > β

Formula Used:
In a right triangle formed by the line of sight and the vertical tower,

​$$\tan(\text{angle}) = \frac{\text{Opposite (height of tower)}}{\text{Adjacent (horizontal distance from foot)}}$$​​

Solution: 

Let the distances of the two cars from the tower be x and x + 120, since one car is farther than the other by 120 m. 

From the tangent definitions:

​$$\tan\alpha = \frac{h}{x}, \tan\beta = \frac{h}{x + 120}$$​​

From $$\tan\alpha = \sqrt{3}$$​,

​$$\frac{h}{x} = \sqrt{3}$$​​

​$$h = x\sqrt{3} \tag{1}$$​​

From $$\tan\beta = \frac{1}{\sqrt{3}},$$​​

​$$\frac{h}{x + 120} = \frac{1}{\sqrt{3}} \\ \ \\ h = \frac{x + 120}{\sqrt{3}} \tag{2}$$​​

Equating equations (1) and (2):

​$$x\sqrt{3} = \frac{x + 120}{\sqrt{3}} \\ \ \\ 3x = x + 120 \\ \ \\ 2x = 120 \\ \ \\ x = 60$$​​

Substitute in equation (1):

​$$h = 60 \times \sqrt{3} = 60\sqrt{3} m$$​​