Question

In the given figure, AB is tower of height 50 m. A man standing on its top, observes two car on the opposite sides of the tower with angles of depression 30° and 45° respectively. Find the distance between the two cars.

Answer 1

Clearly ∠ADB = 45° and ∠ACB = 30° ....[∵ Angle of depression = Angle of elevation]

Now, In ΔABD, we have

tan 45° = `(AB)/(BD)`

⇒ 1 = `50/(BD)`

⇒ BD = 50 m  ...(i)

Also, In ΔABC, we have

tan 30° = `(AB)/(BC)`

⇒ `1/sqrt(3) = (AB)/(BC)`

⇒ BC = `ABsqrt(3) = 50sqrt(3)` m  ...(ii)

From equations (i) and (ii), we get

CD = BC + BD

= `(50sqrt(3) + 50)`m

= `50(sqrt(3) + 1)`m

= 50(1.732 + 1)

= 50 × 2.732

= 136.6 m

Thus, the distance between two cars is 136.6 m.