Question

If the bisector of $\angle A$ of triangle $ABC$ makes an angle $\theta$ with $BC$ then $\sin \theta$ is equal to

• A. $\cos \left(\frac{B-C}{2}\right)$
• B. $\sin \left(\frac{B-C}{2}\right)$
• C. $\sin (B-\frac{A}{2})$
• D. $\sin (C-\frac{A}{2})$

Answer 1

The correct option is A $\cos \left(\frac{B-C}{2}\right)$

$\theta =B+\frac{A}{2}$
We know that in any triangle, $\angle A+\angle B+\angle C=\pi$
$\Rightarrow \theta =B+\frac{\pi -(B+C)}{2}$
$=B+\frac{\pi }{2}-\frac{B}{2}-\frac{C}{2}$
$=\frac{\pi }{2}+\frac{B}{2}-\frac{C}{2}$
$=\frac{\pi }{2}+\frac{B-C}{2}$
$\therefore \sin \theta =\sin (\frac{\pi }{2}+\frac{B-C}{2})=\cos \left(\frac{B-C}{2}\right)$