Question

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

Answer 1

Let AB be the building of height h and AC, the horizontal ground through C, the foot of the building.

Since, the building subtends an angle of 60° at C, hence ∠ACB = 30°.

Let CD be the tower of height 30° m such that ∠CAB = 60°.

From Δs BAC and DCA, we have

$\frac{AC}{AB}$ = cot 30°

⇒ $\frac{AC}{h} = \sqrt{3}$

⇒ AC = $\sqrt{3} \ h$      .............(1)

and, $\frac{DC}{AC}$ = tan 60°

⇒ $\frac{30}{AC} = \sqrt{3}$

⇒ AC = $\frac{30}{\sqrt{3}}$     .............(2)

Equating the values of AC from (1) and (2), we get

$\sqrt{3} \ h = \frac{30}{\sqrt{3}}$

⇒ h = $\frac{30}{\sqrt{3}} \times\frac{1}{\sqrt{3}} =\frac{30}{3} = 10$

Hence, the height of the building is 10 m.