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Question

In ABC, sin2A+sin2B+sin2C=2, then the triangle is always

A
an isosceles triangle
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B
a right angle triangle
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C
an obtuse angle triangle
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D
an acute angle triangle
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Solution

The correct option is B a right angle triangle
sin2A+sin2B+sin2C=21cos2A2+1cos2B2+1cos2C2=2cos2A+cos2B+cos2C=12cos(A+B)cos(AB)=2cos2C2cos(πC)cos(AB)=2cos2C[A+B+C=π]2cosC[cosCcos(AB)]=0[cos(πC)=cosC]2cosC[cos(A+B)+cos(AB)]=0[cos(π(A+B))=cosC]4cosAcosBcosC=0cosAcosBcosC=0
Therefore, one of them should be equal to zero.
So, exactly one of angle is 90

Hence, the triangle is always a right angle triangle.

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