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Question

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 acsin2C+casin2A is


A

12

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B

32

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C

1

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D

3

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Solution

The correct option is D

3


Explanation for Correct Option:

Step 1: Apply the Arithmetic formula

Since A,BandC are in arithmetic progression

B=A+C2

2B=A+C.....(i)

Sum of the angle of a triangle

A+B+C=180°.....(ii)

Step 1: On Solving equations (i) and (ii)

B=60°

By sine rule asinA=bsinB=csinC=k

acsin2C+casin2A=sinAsinC2sinCcosC+sinCsinA2sinAcosA=2sinASinCcosCsinC+2sinCSinAcosAsinA=2sinAcosC+2sinCcosA2sin(A+C)=2sinB=2sin60°=2×32=3

Hence option (4) is the Correct Option


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