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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°. Find the height of the tower. - CBSE Class 10 - Mathematics

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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 with of the canal.

Solution 1

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.

 

 

 

Solution 2

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.

  Is there an error in this question or solution?

APPEARS IN

 NCERT Solution for Mathematics Textbook for Class 10 (2019 to Current)
Chapter 9: Some Applications of Trigonometry
Ex. 9.10 | Q: 11 | Page no. 204
Solution 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°. Find the height of the tower. Concept: Heights and Distances.
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