Question

In a $△ABC$ if the sides are $a=3,b=5$ and $c=5$, then $\sin \frac{B}{2}+\cos \frac{B}{2}$ is equal to

• A. $\sqrt{2}$
• B. $\frac{\sqrt{3}+1}{2}$
• C. $\frac{\sqrt{3}-1}{2}$
• D. $1$

Answer 1

The correct option is A $\sqrt{2}$

We know,
$\cos B=\frac{a^2+c^2-b^2}{2ac}$
$\therefore \cos B=\frac{3^2+4^2-5^2}{2\left(3\right)\left(4\right)}=\frac{9+16-25}{2\left(3\right)\left(4\right)}=0$
$\Rightarrow B={90}^{\circ }$
$\therefore \sin \frac{B}{2}+\cos \frac{B}{2}=\sin {45}^{\circ }+\cos {45}^{\circ }$
$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$