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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.
Solution
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.
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