Question

If $a,b,c$ denote the lengths of the sides of a triangle opposite angles $A,B,C$ of a triangle $ABC$, then the correct relation among $a,b,c,A,B$ and $C$ is given by

• A. $(b+c)\sin \left(\frac{B+C}{2}\right)=a\cos \left(\frac{A}{2}\right)$
• B. $(b-c)\cos \left(\frac{A}{2}\right)=a\sin \left(\frac{B-C}{2}\right)$
• C. $(b-c)\sin \left(\frac{B-C}{2}\right)=a\cos \left(\frac{A}{2}\right)$
• D. $(b-c)\cos \left(\frac{A}{2}\right)=2a\sin \left(\frac{B-C}{2}\right)$

Answer 1

The correct option is B $(b-c)\cos \left(\frac{A}{2}\right)=a\sin \left(\frac{B-C}{2}\right)$

Take $b=k\sin B;c=k\sin C;a=k\sin A$
$\because \frac{b-c}{a}=\frac{k(\sin B-\sin C)}{k\sin A}$
Using tranformation angle formulae, we get
$\frac{k(\sin B-\sin C)}{k\sin A}=\frac{2\cos \left(\frac{B+C}{2}\right)\sin \left(\frac{B-C}{2}\right)}{2\sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)}$
We know that $A+B+C={180}^0$
$\Rightarrow \frac{B+C}{2}={90}^0-\frac{A}{2}$
$\Rightarrow \cos \left(\frac{B+C}{2}\right)=\cos ({90}^0-\frac{A}{2})=\sin \left(\frac{A}{2}\right)$
Substituting $\cos \left(\frac{B+C}{2}\right)=\sin \left(\frac{A}{2}\right)$ we get
$\therefore \frac{b-c}{a}=\frac{2\sin \frac{A}{2}\sin \left(\frac{B-C}{2}\right)}{2\sin \frac{A}{2}\cos \frac{A}{2}}$
$=\frac{\sin \left(\frac{B-C}{2}\right)}{\cos \frac{A}{2}}$
$\therefore (b-c)\cos \frac{A}{2}=\sin \left(\frac{B-C}{2}\right)$