Question

In a $△ABC$ 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

Answer: (A)

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