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 tower is `30^(@)` (see fig.) . FInd the height of the tower and the width of the canal.

Answer

`10sqrt(3) m, 10 m`

Explanation

Let AB be the tower and AC be the central .
`x/h= cot 60^(@) = 1/(sqrt(3)) rArr x = (h)/(sqrt(3)) "………"(i)`
`((x+20))/(h) = cot 30^(@) = sqrt(3) rArr x = (hsqrt(3) - 20)"……."(ii)`
From (i) and (ii) , we get `h/(sqrt(3)) = (hsqrt(3) - 20)`.
`:. h= (3h - 20sqrt(3)) rArr 2 h = 20sqrt(3)rArr h = 10sqrt(3)`.
Putting `h = 10sqrt(3)` in (i), we get ` m = 10 m` (width of the canal).