Question

In a triangle $PQR, A, B$ and $C$ are the angles opposite to the corresponding sides of lengths $a, b$ and $c$ respectively. If the sides are $a = 5$ , $b = 13$ and $c = 12$ then $sin \frac{B}{2} + cos \frac{B}{2} =$

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

Answer 1

Answer: (C)

$\sqrt{2}$

Answer 2

The correct option is(C): $\sqrt{2}$.

We have, $a=5, b=13, c=12$
Here, $a^{2}+c^{2}=5^{2}+12^{2}$
$=169=13^{2}=b^{2}$
$\therefore \Delta\, PQR$ is right triangle right angled at $Q$
$\therefore B=90^{\circ}$
Now, sin $\frac{B}{2}+cos \frac{B}{2}$
$=sin\, 45^{\circ} +cos\, 45^{\circ}$
$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}$
$=\sqrt{2}$