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Question

If the area of the triangle ABC is Δ, such that b2sin2C+c2sin2B=kΔ, then the value of k is

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Solution

b2sin2C+c2sin2B
=b22sinCcosC+c22sinBcosB
=2bcosCbsinC+2ccosBcsinB
=2bcosCcsinB+2ccosBcsinB
[bsinB=csinCbsinC=csinB ]
=2csinB(bcosC+ccosB)
=2csinBa
=412acsinB
=4Δ
k=4

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