The shadow of a pole of height `(sqrt3+1)` metres standing on the ground is found is found to be 2 metres longer when the elevation is `30^@` than when elevation was `alpha`.Then , `alpha`=

Options

A. `75^@`
B. `60^@`
C. `45^@`
D. `30^@`

Answer

C

Explanation

We have,
`OP=(sqrt3+1)m` and AB=2 metres

In `triangle'sAOP` and BOP, we have
tan `30^@=(sqrt3+1)/(OA)` and `tan alpha=(sqrt3+1)/(OB)`
`rArr OA=(sqrt3+1)sqrt3` and OB=`(sqrt3+1)/cot alpha`
`rArr OA-OB=(3+sqrt3)-(sqrt3+1) cot alpha`
`rArr 2=3+sqrt3-(sqrt3+1) cot alpha`
`rArr cot alpha =1 rArr alpha =45^@`