Question

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

• A. $\cos \frac{\angle B-\angle C}{2}$
• B. $\sin \frac{\angle B-\angle C}{2}$
• C. $\sin \frac{\angle B-\angle A}{2}$
• D. None of these

Answer 1

The correct option is A $\cos \frac{\angle B-\angle C}{2}$

In $△ABC$, we have

$\angle A+\angle B+\angle C=\pi \angle A=\pi -\angle B+\angle C$ ..... $\left(i\right)$

In $△ADC$, we have
$\angle ADC+\angle DAC+\angle C=\pi$ ...... $\left(ii\right)$

As $AD$ is the bisector of angle $\angle A$

Substituting in $\left(ii\right)$, we get

$\theta +\frac{\pi -\angle B-\angle C}{2}+\angle C=\pi$

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

So, option A is correct.