Question

Solve.
In A PQR, A Q is an acute angle. Show that PR2 <PQ2+ + QR2?​

Answer 1

Let the lengths of opposing sides be

• $p=\overline{QR}$
• $q=\overline{RP}$
• $r=\overline{PQ}$

By the law of cosines, where $\cos{Q}>0$,

$p^{2}=q^{2}+r^{2}-2qr\cos Q$.

Hence, $p^{2}<q^{2}+r^{2}$.

Answer 2

Correct answer:-

Let,

• $\overline{QR}=p$.
• $\overline{PR}=q$.
• $\overline{PQ}=r$.

Then,

$q^{2}=p^{2}+r^{2}-2pr\cos{Q}$, where $\cos{Q}>0$.

So,

$q^{2}=p^{2}+r^{2}-2pr\cos{Q}<p^{2}+r^{2}$

$\implies q^{2}<p^{2}+r^{2}$

Hence proven that $\overline{PR}^{2}<\overline{PQ}^{2}+\overline{QR}^{2}$.