Exam

JEE Advanced

Year

2009

Paper

Paper 1

Subject

Mathematics

Chapter

Properties of Triangles

JEE Advanced 2009 · Paper 1 · Mathematics · Properties of Triangles

In a A B C with fixed base B C , the vertex A moves such that B+...

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

Step-by-step solution

Given, $\cos B+\cos C=4 \sin ^2 \frac{A}{2}$$\begin{aligned}& \Rightarrow 2 \cos \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right)=4 \sin ^2 \frac{A}{2}\\ & \Rightarrow 2 \sin \frac{A}{2}\left[\cos \left(\frac{B-C}{2}\right)-2 \sin \frac{A}{2}\right]=0 \\ & \Rightarrow \cos \left(\frac{B-C}{2}\right)-2 \cos \left(\frac{B+C}{2}\right)=0 \\ & \Rightarrow-\cos \frac{B}{2} \cos \frac{C}{2}+3 \sin \frac{B}{2} \sin \frac{C}{2}=0 \\ & \Rightarrow \quad \tan \frac{B}{2} \tan \frac{C}{2}=\frac{1}{3}\\ & \Rightarrow \quad \sqrt{\frac{(s-a)(s-c)}{s(s-b)} \cdot \frac{(s-b)(s-a)}{s(s-c)}}=\frac{1}{3}\\ & \Rightarrow \quad \frac{s-a}{s}=\frac{1}{3} \Rightarrow2 s=3 a \\ & \Rightarrow \quad b+c=2 a \end{aligned}$$\\$\therefore$Locus of$A is an ellipse.