The angle of elevation of the top of a building from the foot of a tower is 30\degree and the angle of elevation of the top of the tower from the foot of the building is 60\degree. If the tower is 60m high, find the height of the building.

Step 1: According to the given information construct an imaginary figure.
Given height of tower CD = 60 m
Let the height of the building, AB = h
In right angled triangle BDC,
The angle of elevation of the top of the building from the foot of the tower \angle ADB = 30\degree
The angle of elevation of the top of the tower from the foot of the building \angle CBD = 60\degree
Step 2: Find the height of the building use the suitable trigonometric ratios
From right triangle BCD
\tan \theta = \frac{opposite}{adjacent}
\tan 60\degree = \frac{CD}{BD}
\tan 60\degree = \frac{60}{BD}
\sqrt{3} = \frac{60}{BD}
BD = \frac{60}{\sqrt{3}}
BD = \frac{60}{\sqrt{3}} *\frac{\sqrt{3}}{\sqrt{3}}
BD = \frac{60\sqrt{3}}{3}
BD = 20\sqrt{3}
From right triangle ABD
\tan \theta = \frac{AB}{BD}
\tan 30\degree = \frac{h}{20\sqrt{3}}
\frac{1}{\sqrt{3}} = \frac{h}{20\sqrt{3}}
h = \frac{20\sqrt{3}}{\sqrt{3}}
h = 20 m
Thus height of the building is 20 m.