Question

In a triangle $A B C$ with fixed base $B C$, 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. $b+c=4 a$
• B. $b+c=2 a$
• C. locus of point $A$ is an ellipse
• D. locus of point $A$ is a pair of straight lines

Answer 1

Answer: (B), (C)

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