Question

In $\Delta ABC$, if the sides are $a=3,b=5$ and $c=4$, 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 $\sqrt{2}$

$\cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{3^2+4^2-5^2}{2\times 3\times 4}=0$
As $\cos B=0\Rightarrow B=\frac{\pi }{2}$
$\therefore \sin \frac{B}{2}+\cos \frac{B}{2}=\sin \frac{\pi }{4}+\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}\times 2=\sqrt{2}$