Properties Of Triangles Ques 6
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
(2009)
(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 line
Show Answer
Answer:
Correct Answer: 6.(b,c)
Solution:
Formula:
Trigonometric Functions of Half Angles:
- Given, $\cos B+\cos C=4 \sin ^{2} \frac{A}{2}$
$\Rightarrow 2 \cos (\frac{B+C}{2} ) \cos (\frac{B-C}{2})=4 \sin ^{2} \frac{A}{2}$
$\Rightarrow 2 \sin \frac{A}{2} \cos [(\frac{B-C}{2})-2 \sin \frac{A}{2}]=0$
$\Rightarrow \quad \cos (\frac{B-C}{2})-2 \cos (\frac{B+C}{2})=0$ as $\sin \frac{A}{2} \neq 0$
$\Rightarrow \quad-\cos \frac{B}{2} \cos \frac{C}{2}+3 \sin \frac{B}{2} \sin \frac{C}{2}=0$
$\Rightarrow \quad \tan \frac{B}{2} \tan \frac{C}{2}=\frac{1}{3}$
$\Rightarrow \quad \sqrt{\frac{(s-a)(s-c)}{s(s-b)} \cdot \frac{(s-b)(s-a)}{s(s-c)}}=\frac{1}{3}$
$\Rightarrow \quad \frac{s-a}{s}=\frac{1}{3} \Rightarrow 2 s=3 a$
$\Rightarrow \quad b+c=2 a$
$\therefore$ Locus of $A$ is an ellipse.
BETA
