Question

In $△ABC$, if $a=3,b=5,c=4$, then find value of $\sin \frac{B}{2}+\cos \frac{B}{2}$.

Answer 1

Answer: 1.41

Given, $a=3,b=5,c=4$
$\sin \frac{B}{2}+\cos \frac{B}{2}$
$=\sqrt{\frac{(s-a)(s-c)}{ca}}+\sqrt{\frac{s(s-b)}{ca}}$
$=\sqrt{\frac{(6-3)(6-4)}{4\times 3}}+\sqrt{\frac{6(6-5)}{4\times 3}}$
$=\sqrt{\frac{3\times 2}{4\times 3}}+\sqrt{\frac{6\times 1}{4\times 3}}$
$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$