Question

The angular elevation of a tower from a point is 30°, at a point in a horizontal line to the foot of the tower and 100 metres nearer it is 60°, find the height of the tower. Find also the distance of the first point from the tower.

Answer 1

Let AB represent the tower and P, Q the two points of observation. It is given that PQ = 100 m. Angles of elevation at P and Q are given to be 30° and 60° respectively. Let the required height of the tower be h metres. Let AQ = x metres. Then, 

From right ∆BAQ, \(\frac{h}{x}\) = tan 60° ⇒ h = x tan 60°          …(i) 

From right ∆BAP, \(\frac{h}{x+100}\) = tan 30° ⇒ h = (\(x\) + 100) tan 30°    …(ii) 

Eq. (i) and (ii) ⇒ \(x\) tan 60° = (\(x\) + 100) tan 30°

\(x\sqrt3\) = \(\frac{x+100}{\sqrt3}\) ⇒ 3\(x\) = \(x\) + 100 ⇒ 2\(x\) = 100 ⇒ \(x\) = 50 

∴ From (i) h = \(x\) tan 60° = 50√3 = 50 × 1.732 = 86.6 m (nearly) 

Also, AP = \(x\) + 100 = 50 + 100 = 150 m.