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Question

In a triangle ABC, angles A,B,C are in A.P.Then
limxc(34sinAsinC)|AC|

A
1
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B
2
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C
3
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D
4
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Solution

The correct option is A 1
A,B,C are in A.P
2B=A+C and A+B+C=1800
(A+C)+B=1800
2B+B=3B=1800 or B=600
Using cosine rule,
cosB=cos60012=a2+c2b22ac
a2+c2b2=ac
a2+c2=b2+ac
(ac)2=b2ac
|ac|=(b2ac)
Using sine Rule,
|sinAsinC|=sin2BsinAsinC
2cos(A+C2)sin(AC2)=34sinAsinC
limxc34sinAsinC|AC|=limxc2sin(AC2)|AC|2×2=|1|=1

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