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

Answer 1

The correct option is B $b+c=4a$

Ans. $\left(b\right)$, $\left(c\right)$.
$2\cos \frac{B+C}{2}\cos \frac{B-C}{2}=4{\sin }^2\frac{A}{2}$
or $=\sin \frac{A}{2}\cos \frac{B-C}{2}=2{\sin }^2\frac{A}{2}$ Cancel $\sin \frac{A}{2}\ne 0$
or $=\cos \frac{B-C}{2}=2\sin \frac{A}{2}=2\cos \frac{B+C}{2}$
$\therefore \frac{\cos \frac{B-C}{2}}{\cos \frac{B+C}{2}}=\frac{2}{1}$ Apply Compo. & Divi.
$\frac{2\sin \left(B/2\right)\sin \left(C/2\right)}{\cos \left(B/2\right)\cos \left(C/2\right)}=\frac{1}{3}$
or $\tan \frac{B}{2}\tan \frac{C}{2}=\frac{1}{3}$
or $\frac{s-a}{s}=\frac{1}{3}$
or $2s=3a$ or $b+c=2a\Rightarrow \left(b\right)$
Also $AC+AB>BC$
Hence locus of point $A$ is an ellipse.