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 length of the sides of the triangle opposite to the angles $A,B$ and $C$ respectively then which of the following is true?

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

Answer 1

Answer: (B), (C)

$b+c=2a$
The locus of $A$ is an ellipse

We know that in triangle $ABC$, $A+B+C=180degrees$

$\therefore B+C=180-A$

$\cos B+\cos C=4{\sin }^2\frac{A}{2}$

$2\cos \frac{B+C}{2}\cos \frac{B-C}{2}=4{\sin }^2\frac{A}{2}$

$\cos \frac{B-C}{2}=2\sin \frac{A}{2}$

$2\cos \frac{A}{2}\cos \frac{B-C}{2}=4\cos \frac{A}{2}\sin \frac{A}{2}$

$2\sin \frac{B+C}{2}\cos \frac{B-C}{2}=2\sin A$

$\sin B+\sin C=2\sin A$

$\Rightarrow b+c=2a$

$\Rightarrow AC+AB=2BC$

Now point $A$ moves in such a way that the sum of its distance from points $B$ and $C$ is constant and equal to $2BC$

So its locus is ellipse.

So options $B$ and $C$ are correct.