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← IIT-JEE 2010 · Paper 1
MathematicsSingle correct+3 / -1Question 29 of 84

IIT-JEE 2010 · Paper 1, Mathematics question 29

If the angles A, B and C of a triangle are in an arithmetic progression and if a, b and c denote the lengths of the sides opposite to A, B and C respectively, then the value of the expression acsin⁡2C+casin⁡2A\dfrac ac\sin2C+\dfrac ca\sin2A is

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Answer and worked solution

Official answer: D

Angles in AP give 2B=A+C2B=A+C and A+B+C=πA+B+C=\pi, so B=π3B=\frac\pi3.

Sine rule. ac=sin⁡Asin⁡C\frac ac=\frac{\sin A}{\sin C}, so
acsin⁡2C+casin⁡2A=2sin⁡Acos⁡C+2sin⁡Ccos⁡A=2sin⁡(A+C)=2sin⁡B=3.\frac ac\sin2C+\frac ca\sin2A=2\sin A\cos C+2\sin C\cos A=2\sin(A+C)=2\sin B=\sqrt3.

(D)

Topics:Properties of trianglesTrigonometric ratios and identities
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