Question

Let $p, q$ and $r$ be the side4s opposite to the angles $P, Q$ and $R$, respectively in a $\Delta P Q R$. Then, $2 p r \sin \left(\frac{P-Q+R}{2}\right)$ 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)

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

Answer 2

In $\triangle P Q R, P+Q+R=180^{\circ}$
$\therefore 2 p r \sin \left(\frac{P-Q+R}{2}\right)$
$=2 p r \sin \frac{180^{\circ}-Q-Q}{2}$
$=2 p r \sin \left(90^{\circ}-Q\right)$
$=2 p r \cos Q$
In $\left.\Delta P Q R \cos Q =\frac{p^{2}+r^{2}-q^{2}}{2 p r}\right)$
$=2 p r\left(\frac{p^{2}+r^{2}-q^{2}}{2 p r}\right)=p^{2}+r^{2}-q^{2}$