A TV tower stands vertically on a bank of a 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 another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the with of the canal.

#### Solution 1

In ΔABC,

AB/BC = tan 60º

`(AB)/(BC) = sqrt3`

`BC = (AB)/sqrt3`

In ΔABD,

AB/BD =tan 30º

`(AB)/(BC+CD) = 1/sqrt3`

`(AB)/((AB)/sqrt3+20) = 1/sqrt3`

`(ABsqrt3)/(AB+20sqrt3) = 1/sqrt3`

`3AB = AB+20sqrt3`

`2AB = 20sqrt3`

`AB = 10sqrt3 m`

`BC = (AB)/sqrt3 = ((10sqrt3)/sqrt3)m = 10 m`

Therefore, the height of the tower is `10sqrt3` m and the width of the canal is 10 m.

#### Solution 2

Let PQ=h m be the height of the TV tower and BQ= x m be the width of the canal.

We have,

AB = 20 m, ∠PAQ = 30° ,∠BQ = x and PQ = h

In ΔPBQ,

` tan 60° = (PQ)/(BQ)`

`⇒ sqrt(3) = h/x`

`⇒ h = x sqrt(3) ` ........................(1)

Again in ΔAPQ,

`tan 30° = (PQ)/(AQ)`

`⇒ 1/sqrt(3) = h/(AB +BQ)`

`⇒ 1/sqrt(3) = (x sqrt(3)) / (20+3)` [ Using (1)]

⇒ 3x = 20+x

⇒3x -x =20

⇒ 2x = 20

`⇒ x=20/2`

⇒ x = 10 m

Substituting x = 10 in (i),we get

`h = 10 sqrt(3)m`

So, the height of the TV tower is 10`sqrt( 3)` mand the width of the canal is 10 m.