Question

Continuing with the previous question, what angle do we have at the end of four arcs?

With B as the centre and any radius more than half of AB, draw another arc as shown. This is the third arc. With A as the centre and the same radius as above, cut the previously drawn arc in Y. This would the fourth arc. Join PY.

What is ∠YPA?

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

Answer 1

The correct option is A

${30}^{\circ }$

From B and A, we draw two arcs which intersect each other at Y. These arcs have the same radius so we have BY=AY,
since PB = PA, and PY = PY(common side)

By SSS congruency criterion, $△$ PBY $\cong$ $△$ PAY

So $\angle$ BPY = $\angle$ YPA = $\frac{1}{2}$ $\angle$ BPA = ${30}^{\circ }$