Question

In triangle $ABC$, internal angle bisector of $\angle A$ makes an angle $\theta$ with side $BC$. Value of $\sin \theta$ is equal to:

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

Answer 1

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

$\sin \theta =\sin (\frac{2B+A}{2})=\sin (\frac{2B+\pi -B-C}{2})=\sin (\frac{\pi }{2}+\frac{B-C}{2})=\cos (\frac{B-C}{2})$