Question

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 width of the canal.

Answer 1

In ΔABC,

`("AB")/("BC")` = tan 60°

`("AB")/("BC") = sqrt3`

`"BC" = ("AB")/sqrt3`

In ΔABD,

`("AB")/("BD") `= tan 30°

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

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

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

`3"AB" = "AB"+20sqrt3`

`2"AB" = 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.

Answer 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)` m and the width of the canal is 10 m.