Question

If in a $∆\mathrm{ABC}$, $a=15,b=36,c=39,$then $\sin \frac{C}{2}$ is equal to

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

Answer 1

The correct option is C

$\frac{1}{\sqrt{2}}$

Explanation for the correct option:

Given, $a=15,b=36,c=39$.

$\begin{array}{rcl}s& =& \frac{a+b+c}{2}\\ & =& \frac{15+36+39}{2}\\ & =& 45\end{array}$

We now that,

$\begin{array}{rcl}\sin \frac{C}{2}& =& \sqrt{\frac{(s-a)(s-b)}{ab}}\\ & =& \sqrt{\frac{\left(45-15\right)(45-36)}{15\times 36}}\\ & =& \sqrt{\frac{30\times 9}{15\times 36}}\\ & =& \sqrt{\frac{1}{2}}\\ & =& \frac{1}{\sqrt{2}}\end{array}$

Hence, the correct option is C.