Question

Two towers stands on a horizontal plane. P and Q, where PQ = 30 m are two points on the line joining their feet. As seen from P, the angle of elevation of the tops of the towers are 30° and 60° but as seen from Q are 60° and 45°. The distance between the towers is equal to

(a) 15 (4 + \(\sqrt3\))

(b) 15 (4 - \(\sqrt3\))

(c) 15 (3 + \(\sqrt3\))

(d) 15 (2 + \(\sqrt3\))

Answer 1

(a) 15 (4 + \(\sqrt3\))

Let the two towers be AB and CD whose heights are H m and h m respectively. 

Given, 

PQ = 30 m.

Also,

∠APB = 60°, ∠AQB = 45°,

∠CPD = 30°, ∠CQD = 60°,

From rt Δs CPD and CQD,

PD = h cot 30°,

QD = h cot 60°

∴ PQ = PD – QD

= h (cot 30° – cot 60°)

⇒ 30 = \(h\left({\sqrt3\,-\,\frac{1}{\sqrt3}}\right)\)

⇒ h = 15\(\sqrt3\) m.

From rt Δs AQB and APB,

BQ = H cot 45°,

BP = H cot 60°

∴ PQ = BQ – BP = H (cot 45° – cot 60°)

⇒ 30 = \(H\left({1\,-\,\frac{1}{\sqrt3}}\right)\)

⇒ H = \(\frac{30\sqrt3}{\sqrt3\,-\,1}\times\,\frac{(\sqrt3\,+\,1)}{(\sqrt3\,+\,1)}\)

= 15\(\sqrt3\) (\(\sqrt3\) + 1) m.

∴ Distance between the towers = BD = BQ + QD

= H cot 45° + h cot 60°

= 15\(\sqrt3\) (\(\sqrt3\) + 1) + 15\(\sqrt3\)\(\times\,\frac{1}{\sqrt3}\)

= 45 +  15\(\sqrt3\)  + 15

= 60 +  15\(\sqrt3\) 

= 15 (4 + \(\sqrt3\)) m.