A vertical pole AB is standing at the centre B of a square PQRS. If PR subtends an angle of 90° at the top, A of the pole, then the angle subtended by a side of the square at A is:

Solution (By Examveda Team)

$$\eqalign{ & PA = AR \cr & \angle APB = \angle ARB = {45^ \circ } \cr & {\text{If }}PR = \sqrt 2 x;\,\,PB = \frac{x}{{\sqrt 2 }} \cr & {\text{From }}\Delta APB \cr & \tan {45^ \circ } = \frac{{AB}}{{PB}} \cr & AB = PB = \frac{x}{{\sqrt 2 }} \cr & \therefore PA = \sqrt {\frac{{{x^2}}}{2} + \frac{{{x^2}}}{2}} = \sqrt {{x^2}} = x \cr & \therefore QA = PQ = PA = x \cr & \therefore \angle PAQ = {60^ \circ } \cr} $$