Question

In a $△$ ABC w ith fixed base BC, the vertex A moves
such that cos B + cos C = 4 $si{n}^2\frac{A}{2}$. If a, b and c denote
th e lengths of th e sides of th e 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 line

Answer 1

Answer: (B), (C)

Given, cos B + cos C = 4 $si{n}^2\frac{A}{2}$
$\Rightarrow 2cos\left(\frac{B+C}{2}\right)cos\left(\frac{B-C}{2}\right)=4si{n}^2\frac{A}{2}$
$\Rightarrow 2sin\frac{A}{2}\left[cos\right(\frac{B-C}{2})-2sin\frac{A}{2}]=0$
$\Rightarrow cos\left(\frac{B-C}{2}\right)-2cos\left(\frac{B+C}{2}\right)=0$
as sin $\frac{A}{2}\ne 0$
$\Rightarrow -cos\frac{B}{2}cos\frac{C}{2}+3sin\frac{B}{2}sin\frac{C}{2}=0$
$\Rightarrow tan\frac{B}{2}tan\frac{C}{2}=\frac{1}{3}$
$\Rightarrow \sqrt{\frac{(s-a)(s-c)}{s(s-b)}.\frac{(s-b)(s-a)}{s(s-c)}}\frac{1}{3}$
$\Rightarrow \frac{s-a}{s}\frac{1}{3}$
$\Rightarrow 2s=3a$
$\Rightarrow b+c=2a$.
$\therefore$ Locus of A is an ellipse