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Question

If in ABC,cosA+cosB+cosC=3/2, then triangle ABC is

A
right angled
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B
isosceles
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C
acute
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D
equilateral
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Solution

The correct option is D equilateral
cosA+cosB+cosC=3/2
2(2cosA+B2cosAB2)+2(12sin2C2)=3
As A+B+C=π
(4sinC2cosAB2)+(24sin2C2)=3
4sin2C24cos(AB2)sinC2+1=0
Since, D0
16cos2(AB2)4(4)(1)0
cos2(AB2)1
But cos2x[0,1]
So, cos2(AB2)=1
AB2=0A=B
Similarly, B=C
Therefore, A=B=C
Hence, It is an equilateral triangle

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