Question

in triangle PQR, Q is an acute angle. show that PR square < PQ square + QR square

Answer 1

Answer:

The proof is explained below.

Step-by-step explanation:

Given in triangle PQR, Q is an acute angle.

We have to prove that $PR^{2}<PQ^{2}+QR^{2}$

By Cosine law,

$cos\theta=\frac{a^{2}+b^{2}-c^{2}}{2ab}$

⇒ $cosQ=\frac{PQ^{2}+QR^{2}-PR^{2}}{2(PQ)(QR)}$

Here, ∠Q is acute ∴ cos Q > 0

⇒ $\frac{PQ^{2}+QR^{2}-PR^{2}}{2(PQ)(QR)}>0$

⇒ ${PQ^{2}+QR^{2}-PR^{2}} > 0$

⇒  $PR^{2}<PQ^{2}+QR^{2}$