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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
$(B,C)$
$(B,D)$
$(A,C)$
$(A,D)$
Solution
We know that in triangle $ABC , A + B + C =18$ o degrees
$\therefore B + C =180- A$
$\cos B +\cos C =4 \sin ^2 \frac{ A }{2}$
$2 \cos \frac{ B + C }{2} \cos \frac{ B – C }{2}=4 \sin ^2 \frac{ A }{2}$
$\cos \frac{ B – C }{2}=2 \sin \frac{ A }{2}$
$2 \cos \frac{ A }{2} \cos \frac{ B – C }{2}=4 \cos \frac{ A }{2} \sin \frac{ A }{2}$
$2 \sin \frac{ B + C }{2} \cos \frac{ B – C }{2}=2 \sin A$
$\sin B +\sin C =2 \sin A$
$\Rightarrow b + C =2 a$
$\Rightarrow AC + AB =2 BC$
Now point $A$ moves in such a way that the sum of its distance from points $B$ and $C$ is constant and equal to $2$ $BC$
So its locus is ellipse.
So options $B$ and $C$ are correct.
