Question

In $\Delta ABC,$ if $\sin \frac{A}{2}.\cos \frac{B}{2}=\frac{a+c-b}{2c}k$

then $k=?$

• A. $cos\frac{A}{2}$
• B. $sin\frac{B}{2}$
• C. $sin\frac{C}{2}$
• D. $cos\frac{C}{2}$

Answer 1

The correct option is D $cos\frac{C}{2}$

$=\sqrt{\frac{(s-b)(s-c)}{bc}}\times \sqrt{\frac{sc(s-b)}{ac}}$
$=\frac{(s-b)}{c}\sqrt{\frac{s(s-c)}{ba}}$
$=\frac{a+c-b}{2c}cos\left(\frac{c}{2}\right)$
$\therefore k=cos\frac{c}{2}$