From the top of a cliff 90 metre high, the angles of depression of the top and bottom of a tower are observed to be 30° and 60° respectively. The height of the tower is :

Given: CD = 90 meter;
∠KBD = 30° & ∠CAD = 60°
BD × cos30° = BK
AD × cos60° = AC
BD × sin30° = DK
AD × sin60° = DC
BK = AC
BD × cos30° = AD × cos60°
√ 3 × BD = AD ____________ (1)
Now, DC = DK + KC
DC - DK = AB [As, KC = AB]
(AD × sin60°) - (BD × sin30°) = AB
Using equation (1)
($$\sqrt{3}$$ ×BD×$$\frac{\sqrt{3}}{2}$$)−BD2=AB(3×BD×32)−BD2=AB
AB = BD _____________ (2)
Now, AD × sin 60° = DC = 90 meter
AD=$$\frac{180}{\sqrt{3}}$$
From (1)
BD=AD/$$\sqrt{3}$$
=180/$$\sqrt{3} \times \sqrt{3} $$ = 60 meter
From (2) AB = 60 meter
Option B is the correct answer.