The angle of elevation of the top of a vertical tower from a point on the ground is `60°`. From another point 10 m vertically above the first, its angle of elevation is `30°`. Find the height of the tower.

Options

A. `5`
B. `10`
C. `15`
D. `20`

Answer

C

Explanation

Let the height of the vertical tower, OT=H
and OP=AB=xm
Given that, AP=10m
and `angleTPO=60^(@), angleTAB=45^(@)`
Now, in `DeltaTPO, tan60^(@)= (OT)/(OP) = H/x`
`rArr sqrt(3)=H/x`
`rArr x=H/sqrt(3)`
and in `DeltaTAB`, `tan45^(@)= (TB)/(AB) = (H-10)/(x)`
`rArr 1=(H-10)/(x) rArr x=H-10`
`rArr H/sqrt(3) = H-10` [from Eq. (i)]
`rArr H-H/sqrt(3) = 10 rArr H(1-1/sqrt(3)) = 10`
`rArr H(sqrt(3)-1)/(sqrt(3)) = 10`
`therefore H= (10sqrt(3))/(sqrt(3))=10`
`therefore H=(10sqrt(3))/(sqrt(3)-1). ((sqrt(3)+1)/(sqrt(3)+1))` [By rationalisation]
`=(10sqrt(3)(sqrt(3)+1))/(3-1) = (10sqrt(3)(sqrt(3)+1))/(2)`
`rArr =5sqrt(3)(sqrt(3)+1)=5(sqrt(3)+3)`m.
Hence, the required height of the tower is `5sqrt(3)+3)` m.