Question

A man on the deck of a ship is 10 m above the water level. He observes that the angle of elevation of the top of a cliff is 45° and the angle of depression of the base is 300. Calculate the distance of the cliff from the ship and the height of the cliff.

Answer 1

Let H be the height of the cliff CE. And a man is standing on the ships at the height of 10 meters above from the water level.

Let AB = 10, BC = x, AD = BC, AB = DC, DE = h. 

∠ACB = 30° and ∠DAE = 45°

We have tofind H and xT

The corresponding figure is as follows

In Δ ABC

`=> tan C = (AB)/(BC)`

`=> tan 30^@ = = 10/x`

`=> 1/sqrt3 = 10/x`

`=> x = 10sqrt3`

Again in ΔDAE

`=> tan A = (DE)/(AD)`

`=> tan 45^@ = h/x`

`=> 1 = h/x`

=> x = h

`=> x = 10sqrt3`

Therefore H = h + 10

`=> H = 10sqrt3 + 10`

`=> H = 10(sqrt3 + 1)`

`=> H = 27.32`

Hence the required distance is `10sqrt3` and height  is 27.32 m