Question

lf in a triangle $ABC$, in the usual notation, $2a\cos \left(\frac{B-C}{2}\right)=b+c$ and $B\ne C$, then the measure of the angle $A$ is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is A $\frac{\pi }{3}$

$\cos \left(\frac{B-C}{2}\right)=\frac{b+C}{2a}$
$\cos \left(\frac{B-C}{2}\right)=\frac{\sin B+\sin C}{2\sin A}$
$\cos \left(\frac{B-C}{2}\right)=\frac{2\sin \left(\frac{B+C}{2}\right)\cos \left(\frac{B-C}{2}\right)}{2\times 2\sin A/2\cos A/2}$
$2sinA/2\cdot \cos \left(\frac{\pi -(B+C)}{2}\right)=\sin \left(\frac{B+C}{2}\right)$
$2sin\left(A/2\right)\cdot sin\left(\frac{B+C}{2}\right)=sin\left(\frac{B+C}{2}\right)$
$\sin A/2=1/2$
$A/2=\pi /6$
$A=\pi /3$