Question

A man from the top of a 100 metres high tower sees a car moving towards the tower at an angle of depression of 30°. After sometime, the angle of depression becomes 60°. The distance (in metres) travelled by the car during this time is

(a) 200√3

(b) \(\frac{100}{\sqrt3}\)

(c) \(\frac{200}{\sqrt3}\)

(d) 100√3

Answer 1

(c) \(\frac{200}{\sqrt3}\)m

Let PQ be the given tower of height 100 metres. 

Let P be the position of the man and R and S the positions of the car moving towards the tower. Given angle of depression at points R and S are 30° and 60° respectively. 

Distance travelled by the car = RS = QR – QS.           ...(i) 

In rt. Δ PQR, tan 30° = \(\frac{PQ}{QR}=\frac{100}{QR}\)

⇒ \(\frac{1}{\sqrt3}\) = \(\frac{100}{QR}\) ⇒ QR = 100√3 m                         .....(ii)

In rt. Δ PSQ, tan 60° = \(\frac{PQ}{QS}=\frac{100}{QS}\)

⇒ √3 = \(\frac{100}{QS}\) ⇒ QS = \(\frac{100}{\sqrt3}\)m                     .....(iii)

∴ From (i), (ii) and (iii)

RS = 100√3 - \(\frac{100}{\sqrt3}\)=  100√3 - \(\frac{200\sqrt3}{3}\) = \(\frac{200}{\sqrt3}\) m.