Question

In a triangle $ABC$ with fixed base $BC$, the vertex $A$ moves such that
$\cos B+\cos C=4{\sin }^2\frac{A}{2}$
If $a,b$ and $c$ denote the lengths of the sides of the triangle opposite to the angles $A,B$ and $C$, respectively, then

• A. $b+c=4a$
• B. $b+c=2a$
• C. locus of point $A$ is an ellipse
• D. locus of point $A$ is a pair of straight lines

Answer 1

Answer: (B), (C)

$2\cos \left(\frac{B+C}{2}\right)\cos \left(\frac{B-C}{2}\right)=4{\sin }^2\frac{A}{2}$
$\cos \left(\frac{B-C}{2}\right)=2\sin (A/2)$
$\Rightarrow \frac{\cos \left(\frac{B-C}{2}\right)}{\sin A/2}=2$
$\Rightarrow \frac{\sin B+\sin C}{\sin A}=2$
$\Rightarrow b+c=2a$ (constant)