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Question

If In Δ ABC the sides opposite to angles A,B,C are denoted by a,b,c respectively, and the sides a,b,c are in A.P.

then?

A
B>60
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B
B<60
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C
B60
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D
B=AC
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Solution

The correct option is C B60
Given that, 2b=a+c
From sine rule: 2sinB=sinA+sinC
4sinB2cosB2=2sin(A+C2)cos(AC2)
2sin(π2A+C2)cos(π2A+C2)=sin(A+C2)cos(AC2)
2cos(A+C2)=cos(AC2)
cos(A+C2)cos(AC2)=12
By componendo-dividendo
cos(A+C2)+cos(AC2)cos(A+C2)cos(AC2)=1+212
cosA2cosC2sinA2sinC2=3
tanA2tanC2=13 ...(1)
As A.MG.M
Therefore, tanA2+tanC22tanA2tanC2=13 ...(2)

In ΔABC: B2=π2(A+C2)
cotB2=tan(A+C2)=tanA2+tanC21tanA2tanC2
cotB22/31(1/3)=3 [ from (1) and (2) ]
Therefore, B60
Ans: C

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