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Question

lf Sin2A+Cos2B+Sin2C=1, then the triangle ABC is

A
isosceles
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B
equilateral
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C
right angled
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D
scalene
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Solution

The correct option is C right angled
sin2A+cos2B=cos2C
sin2A=(cosCcosB)(cosC+cosB)
sin2A=2×2cos(C+B2)cos(CB)2sin(B+C2)sin(BC2)
sin2A=sin(BC)sin(B+C)
sinA=sin(BC)
A=BC
A+B+C=π
2B=π
B=π2

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