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:

Answer: Option C

Solution

Answer: Option C

Solution:

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