The angle of depression of the bottom of the top of 20 meter high building, front he too if the tower is 30° and the angle of elevation of the top of tower from the foot of building is 60°. Find the height of the tower.

Solution:

Given,

Height of the building (AD) = 20m

Angle of depression ( angle.FEA ) = 30°

Angle of elevation (angle.CDE ) = 60°

From the figure,

ABCD is a rectangle. 

So, AD = BC = 20m and AB = CD.

In right angled ∆CDE

tan 60° = $\dfrac{ EB + BC}{CD}$

or, √3 = $\dfrac{ x + 20}{CD}

or, √3 CD = x + 20

or, CD = $\dfrac{x +20}{√3}$

In right angled ∆ABE

tan30° = $\dfrac{BE}{AB}$

or, $\dfrac{1}{√3}$ = $\dfrac{x}{CD}$

or, CD = √3 x

or, $\dfrac{x+20}{√3} $ = √3x

or, x + 20 = √3 * √3x

or, x + 20 = 3x

or, 3x - x = 20

or, 2x = 2*10

So, x = 10 meter

Now,

Height of tower (CE) = BE + BC

= x + 20m

= 10m + 20m

= 30m

Therefore, the required height of the building is 30 meters.