Correct option is (A) P = Q
In \(\triangle P Q R, \angle R=90^{\circ}\)
\(\sin P=\frac{Q R}{P Q} \text{ and } \sin Q=\frac{P R}{P Q}\)
As
\( \sin P=\sin Q\)
\( \frac{Q R}{P Q}=\frac{P R}{P Q} \)
\(Q R=P R\)
So, \(\angle P=\angle Q\) [\(\because\) Angles opposite to equal sides are equal.]
