Question

A vertical pole PO is standing at the centre O of a square ABCD. If AC subtends an angle of ${90}^0$ at the point P of the pole, then the angle subtended by a side of the square at the point P is

• A. ${35}^0$
• B. ${45}^0$
• C. ${30}^0$
• D. ${60}^0$
• E. None of the above

Answer 1

Answer: (D)

${60}^0$

Given $\angle APO+\angle CPO={90}^0$
$\Rightarrow \angle APO=\angle CPO={45}^0$
$=\angle BPO=\angle DPO$
$\cos \angle APB=\cos \theta =\frac{A{P}^2+B{P}^2-A{B}^2}{2AP\cdot BP}$
$=\frac{A{O}^2+O{P}^2+B{O}^2+B{P}^2-A{B}^2}{2AP\cdot BP}$
$=\frac{O{P}^2}{AP\cdot BP}=\frac{OP}{AP}\cdot \frac{OP}{BP}$
$=\sin {45}^0\times \sin {45}^0=\frac{1}{2}$