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← IIT-JEE 2009 · Paper 1
MathematicsMultiple correct+4 / -1Question 31 of 60

IIT-JEE 2009 · Paper 1, Mathematics question 31

In a triangle ABCABC with fixed base BCBC, the vertex AA moves such that
cos⁡B+cos⁡C=4sin⁡2A2.\cos B+\cos C=4\sin^2\frac A2.
If aa, bb and cc denote the lengths of the sides of the triangle opposite to the angles AA, BB and CC, respectively, then

Choose one or more options
Answer and worked solution

Official answer: B, C

Transform. cos⁡B+cos⁡C=2cos⁡B+C2cos⁡B−C2=2sin⁡A2cos⁡B−C2\cos B+\cos C=2\cos\frac{B+C}2\cos\frac{B-C}2=2\sin\frac A2\cos\frac{B-C}2. The condition becomes
cos⁡B−C2=2sin⁡A2.\cos\frac{B-C}2=2\sin\frac A2.

Sides. By the sine rule, b+ca=sin⁡B+sin⁡Csin⁡A=cos⁡B−C2sin⁡A2=2\dfrac{b+c}a=\dfrac{\sin B+\sin C}{\sin A}=\dfrac{\cos\frac{B-C}2}{\sin\frac A2}=2. So b+c=2ab+c=2a.

Locus. AB+AC=2 BCAB+AC=2\,BC, a constant larger than BC: A moves on an ellipse with foci B and C.

(B), (C)

Topics:EllipseTrigonometric ratios and identities
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