# How do you find the exact value of the third side given triangle PQR, p=6, q=sqrt2 and mangleR=pi/4?

Dec 23, 2016

$r = \sqrt{26} = 5.1$

#### Explanation:

We can find the value of third side if two sides and measure of included angle between them is given. We use Law of cosines for this, according to which if two sides $p$ and $q$ and included angle $R$ are given, then

${r}^{2} = {p}^{2} + {q}^{2} - 2 p q \cos R$

Here, we are given $p = 6$, $q = \sqrt{2}$ and $R = \frac{\pi}{4}$

Hence, ${r}^{2} = {6}^{2} + {\left(\sqrt{2}\right)}^{2} - 2 \times 6 \times \sqrt{2} \times \cos \left(\frac{\pi}{4}\right)$

= $36 + 2 - 12 \times \sqrt{2} \times \frac{1}{\sqrt{2}} = 38 - 12 = 26$

Hence $r = \sqrt{26} = 5.1$