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A triangle A B C with fixed base B C , t...

A triangle `A B C` with fixed base `B C` , the vertex `A` moves such that `cosB+cosC=4sin^2A/2dot` If `a ,ba n dc ,` denote the length of the sides of the triangle opposite to the angles `A , B ,a n dC` , respectively, then (a)`b+c=4a` (b) `b+c=2a` (c)the locus of point `A` is an ellipse (d)the locus of point `A` is a pair of straight lines

A

b+c=4a

B

b+c=2a

C

the locus of point A is an ellipse

D

the locus of point A is a pair of straight lines

Text Solution

Verified by Experts

Given `cos B+cos C=4 sin^(2).(A)/(2)`
`or 2 cos((B+C)/(2))cos ((B-C)/(2))=4 sin^(2).(A)/(2)`
or `cos.((B-C)/(2))=2sin.((A)/(2))`
`or 2 sin.(B+C)/(2)cos.(B-C)/(2)=4 sin.(A)/(2)cos.(A)/(2)`
or `sin B+sin C=2 sin A`
or `b+c=2a`
Thus, sum of two variable sides is constant .
Hence the locus of vertex A is an ellipse with B and C as foci,
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