The angular elevation of a tower OP at a point A due south of it is `60^@` and at a point B due west of A, the elevation is `30^@`. If AB=3m, the height of the tower is

Options

A. `2sqrt3`m
B. `2sqrt6`m
C. `(3sqrt3)/2` m
D. `(3sqrt6)/4` m

Answer

D

Explanation

Let OP be the tower of height h metre.
In `triangle's`AOP and BOP , we have
`tan 60^@=h/(OA)` and `tan 30^@=h/(OB)`
`rArr OA=h/sqrt3` and `OB=sqrt3h`

Now,
`OB^2=OA^2+AB^2`
`rArr 3h^2=h^2/3 + 9 rArr (8h^2)/3 =9 rArr h=(3sqrt3)/(2sqrt2)=(3sqrt6)/4`