Question

A TV tower stands vertically on a bank of a river/canal. 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/canal.

Answer 1

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

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

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 river/canal is 10 m.