The angle of elevation of the top of a vertical tower from a point A, due east of it is `45^(@)`. The angle of elevation of the top of the same tower from a point B, due south of A is `30^(@)`. If the distance between A and B is `54sqrt(2)`m, then the height of the tower (in metres), is :

Options

A. 54
B. 108
C. `27sqrt2`
D. `36sqrt2`

Answer

C

Explanation

Let the height of tower OP be h metres. In right triangles AOP and BOP , we obtain

`tan45^@=h/(OA)` and `tan 30^@=h/(OB)`
`rArr OA=h` and `OB=sqrt3h`
Applying Pythagoras theorem in `triangleAOB`, we obtain
`OA^2+OB^2=AB^2`
`rArr h^2+3h^2 =(54sqrt3)^2`
`rArr 4h^2 =54xx54xx2`
`rArr h^2=27xx27xx2 rArr h=27sqrt2` m