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 the angle $\angle YPA$?

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

Answer 1

The correct option is A

30°

From B and A, we draw two arcs which intersect each other in Y.

These arcs have the same radius so we have BY=AY,
since PB = PA, and PY = PY(common side)

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

Clearly PY is the angle bisector of $\angle BPA$.
$\therefore \angle YPA={30}^{\circ }$