Question

In a $\triangle A B C$ 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)

$\sqrt{2}$

Answer 2

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