Question

The angle of elevation of a stationary cloud from a point 2500 m above a lake is 15° and the angle of depression of its reflection in the lake is 45°. What is the height of the cloud above the lake level? (Use tan 15° = 0.268)

Answer 1

Let AB be the surface of lake and P be the point of observation such that AP = 2500 m. Let C be the position of cloud and C' be the reflection in the lake. Then CB = C'B

Let PQ be the perpendicular from P on CB.

Let PQ = x m, CQ = h, QB = 2500 m. then CB = h + 2500 consequently C'B = h + 2500 m. and ∠CPQ - 15°, .∠QPC' = 45°

Here we have to find the height of cloud.

We use trigonometric ratios.

In ΔPCQ

`=> tan 15° = (CQ)/(PQ)`

`=> 2 - sqrt3 = h/x`

`=> x = h/(2 - sqrt3)`

Again in Δ PQC

`=> tan 45^@ = (QB + BC')/(PQ)`

`=> 1 = (2500 + h + 2500)//x`

`=> x = 5000 } h`

`=> h /()2 - sqrt3 = 5000 + h`

`=> h = 2500 (sqrt3 - 1)`

`=> CB = 2500 + 2500 (sqrt3 - 1)`

`=> CB = 2500sqrt3`

Hence the height of cloud is `2500sqrt3` m