#### Question

A T.V. Tower stands vertically on a bank of a river. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From a point 20 m away this point on the same bank, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the river.

#### Solution

Let *AB* be the T.V tower of height hm on a bank of river and *D* be the point on the opposite of the river. An angle of elevation at top of the tower is 60° and from a point 20m away then an angle of elevation of the tower at the same point is 30°. Let *AB = h* and *BC = x.*

Here we have to find height and width of the river.

The corresponding figure is here

In ΔCAB

`=> tan 60^@ = (AB)/(BC)`

`=> sqrt3 = h/x`

`=> sqrt3x = h`

`=> x = h/sqrt3`

Again in ΔDBA

`=> tan 30^@ = (AB)/(BC)`

`=> 1/sqrt3 = h/(20 + x)`

`=> sqrt3h = 20 + x`

`=> sqrt3h = 20 + h/sqrt3`

`=> sqrt3h - h/sqrt3 = 20`

`=> (2h)/sqrt3 = 20`

`=> h = 10sqrt3`

`=> x = (10sqrt3)/sqrt3`

=> x = 10

Hence the height of the tower is `10sqrt3` m and width of rived is 10 m