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Question

In ΔABC, b2+c2b2c2=sin(B+C)sin(BC) then the triangle is

A
A right angled triangle
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B
An isosceles triangle
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C
A right angled isosceles triangle
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D
A scalene triangle
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Solution

The correct option is A A right angled triangle
We have, A+B+C=π

b=KsinB
c=KsinC

Now,
sin2B+sin2Csin2Bsin2C=sinAsin(BC)

sin2B+sin2C(sinBsinC)(sinB+sinC)=sinAsin(BC)
sin2B+sin2C2sin(C+B2)cos(BC2)×2cos(B+C2)×sin(BC2)=sinAsin(BC)

sin2B+sin2Csin(BC)sin(B+C)=sinAsin(BC)
sin2A=sin2B+sin2C
sin2C=cos2Bcos2A
=(cosBcosA)(cosA+cosB)
=2×2cos(B+A2)cos(BC2)sin(A+B2)sin(AB2)
sin2C=sin(A+B)sin(AB)
sin2C=sinCsin(AB)
sinC=sin(AB)
C=AB
A=C+B
A+B+C=π

A=π2

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