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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^2(A/2)dot` 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

locus of point A is an ellipse

D

locus of point A is a pair of straight lines

Text Solution

Verified by Experts

The correct Answer is:
B, C
`cos B + cos C = 4 "sin"^(2) (A)/(2)`
`rArr 2 "cos"(B+C)/(2) = 4 "sin"(A)/(2) " " ( :' (B+C)/(2) = "sin"(A)/(2))`
`= 2 "cos"(A)/(2) "cos"(B-C)/(2) = 4 "sin"(A)/(2) "cos"(A)/(2)`
`rArr 2 "sin"(B+C)/(2) "cos"(B-C)/(2) = 2 sin A`
`rArr sin B + sin C = 2 sin A`
`rArr b + c = 2a` (Using sine rule)
Thus sum of two variable sides b and c is constant `2a`. So locus of vertex A is ellipse with vertices B and C as its foci.
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