In a triangle A B C with fixed base B C , the vertex A moves such that cos B + cos C = 4 sin 2 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 [

Question

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]

Smart Achievers · Accepted Answer

(2, 3)

$In ∆ A B C, \cos B + \cos C = 4 \sin^{2} \frac{A}{2} (Given)$

$\Rightarrow 2 \cos \frac{B + C}{2} \cos \frac{B - C}{2} - 4 \sin^{2} \frac{A}{2} = 0$

$\Rightarrow 2 \sin \frac{A}{2} [\cos \frac{B - C}{2} - 2 \sin \frac{A}{2}] = 0$

$\Rightarrow \sin \frac{A}{2} = 0 or \cos \frac{B - C}{2} - 2 \cos \frac{B + C}{2} = 0$

$Since in a triangle, \sin \frac{A}{2} \neq 0$

$∴$ $\cos \frac{B - C}{2} - 2 \cos \frac{B + C}{2} = 0$ $\Rightarrow \frac{\cos (\frac{B + C}{2})}{\cos (\frac{B - C}{2})} = \frac{1}{2}$

$By componendo and dividendo, we get$

$\frac{\cos (\frac{B + C}{2}) + \cos (\frac{B - C}{2})}{\cos (\frac{B + C}{2}) - \cos (\frac{B - C}{2})} = \frac{1 + 2}{1 - 2} = - 3$

$\Rightarrow \frac{2 \cos \frac{B}{2} \cos \frac{C}{2}}{- 2 \sin \frac{B}{2} \sin \frac{C}{2}} = - 3$ $\Rightarrow \tan \frac{B}{2} \tan \frac{C}{2} = \frac{1}{3}$

$\Rightarrow \sqrt{\frac{(s - a) (s - c)}{s (s - b)}} \sqrt{\frac{(s - a) (s - b)}{s (s - c)}} = \frac{1}{3}$

$\Rightarrow \frac{s - a}{s} = \frac{1}{3} \Rightarrow 2 s = 3 a$ $\Rightarrow a + b + c = 3 a \Rightarrow b + c = 2 a$

$i.e. A C + A B = constant$ $(∵ Base B C = a is given to be constant)$

$∴$ $Locus of A is an ellipse.$

Solution

Q.
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]

1 $b + c = 4 a$

2 $b + c = 2 a$

3 locus of point $A$ is an ellipse

4 locus of point $A$ is a pair of straight lines

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Solution

Q. In a triangle ABC with fixed base BC, the vertex A moves such that cosB+cosC=4sin2A2. 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]

1 b+c=4a

2 b+c=2a

3 locus of point A is an ellipse