A vertical pole PO is standing at the centre O of a square ABCD. If AC subtends an angle of `90^@` at the top , P of the pole, then the angle subtended by a side of the square at P is

Options

A. `45^@`
B. `30^@`
C. `60^@`
D. none of these

Answer

C

Explanation

Since AC subtends a right angle at P.
`therefore angleOPA=angleOPC=pi/4`
Let OP=h.
In `triangleOPA`, we have
tan `pi/4=(OA)/(OP)` and cos `pi/4=(OP)/(PA)`
`rArr OA=OP` and `PA=sqrt2OP`
`rArr OA=h` and `PA=sqrt2h`
`rArr AC=2h` and `PA=sqrt2h`

Also,
`AC=sqrt2AB rArr 2h=sqrt2 AB rArr AB =sqrt2h`
`therefore PA=PB=AB=sqrt2h`
`trianglePAB` is an equilateral triangle
`rArr angleAPB=60^@`