Question

Let $p,q$ and $r$ be the sides opposite to the angles $P,Q$ and $R$ respectively in a $△PQR$. Then $2pr\sin \left(\frac{P-Q+R}{2}\right)$ equals

• A. $p^2+q^2+r^2$
• B. $p^2+r^2-q^2$
• C. $q^2+r^2-p^2$
• D. $p^2+q^2-r^2$

Answer 1

Answer: (B)

$p^2+r^2-q^2$

In $△PQR,P+Q+R={180}^{\circ }$

$\therefore 2pr\sin \left(\frac{P-Q+R}{2}\right)=2pr\sin \frac{{180}^{\circ }-Q-Q}{2}$

$=2pr\sin ({90}^{\circ }-Q)$

$=2pr\cos Q$

$(\text{In}△PQR,\cos Q=\frac{p^2-r^2-q^2}{2pr})$

$=2pr\left(\frac{p^2+r^2-q^2}{2pr}\right)=p^2+r^2-q^2$