Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If the bisector of angle $A$ of triangle $ABC$ makes an angle $\theta$ with $BC$,

then sin $\theta$ is equal to?

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

Answer 1

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

$Now,\frac{\sin \theta }{b}=\frac{\sin \frac{A}{2}}{x}and\frac{\sin (180°-\theta )}{c}=\frac{\sin \frac{A}{2}}{a-x}$

Dividing both equations:

$\frac{c}{b}=\frac{a-x}{x},\frac{a}{x}=\frac{c}{b}+1x=\frac{ab}{b+c}$

$Now,\frac{Sin\theta }{b}=\frac{\sin \frac{A}{2}}{\frac{ab}{b+c}}\sin \theta =\frac{b\sin A/2}{ab}(b+c)\sin \theta =\frac{(b+c)\sin A/2}{a}\Rightarrow \frac{\sin B+\sin C}{2\sin \frac{A}{2}\cos \frac{A}{2}}\times \sin \frac{A}{2}\sin \theta =\frac{2\sin \left(\frac{B+C}{2}\right)\cos \left(\frac{B-C}{2}\right)}{2\cos \left(\frac{A}{2}\right)}\left(SineRule\right)$

$\sin \theta =\cos \left(\frac{B-C}{2}\right)$

(A) is correct.