Question

If $\angle APB={60}^{\circ }$ and with B as the centre and any radius more than half of AB, another arc is drawn as shown. Taking A as the centre and the same radius as above, cut the previously drawn arc in Y and PY is drawn.

What will be the value of $\angle BPY$?

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${90}^{\circ }$
• D. ${180}^{\circ }$

Answer 1

The correct option is A

${30}^{\circ }$

From the given construction, we get PY as the angle bisector of $\angle APB$.
Therefore $\angle BPY=\angle APY=\frac{60}{2}={30}^{\circ }$.