#### Question

At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is 30˚. The angle of depression of the reflection of the cloud in the lake, at A is 60˚.

Find the distance of the cloud from A.

#### Solution

Let AB be the surface of the lake and P be the point of observation such that AP = 20 metres. Let C be the position of the cloud and C’ be its reflection in the lake.

Then CB = C’B. Let PM be perpendicular from P on CB.

Then m∠CPM=30º and m∠C'PM=60°

Let CM = h. Then CB = h + 20 and C’B = h + 20.

In ΔCMP we have,

`tan30^@="CM"/"PM"`

`1/sqrt3=h/"PM"`

`PM=sqrt3h....................(i)`

In ΔPMC' we have,

`tan 60^@="C'M"/"PM"`

`sqrt3="C'B+BM"/"PM"`

`sqrt3=(h+20+20)/"PM"................(ii)`

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

`sqrt3h=(h+20+20)/sqrt3`

3h=h+40

h=20m

Now,CB=CM + MB =h +20= 20+ 20 = 40.

Hence, the height of the cloud from the

surface of the lake is 40 metres.