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Question

Let a,b and c be the sides opposite to angles A, B and C in a triangle ABC. If sinA(sinA+cosBsinB)+cosA(cosA+sinB+cosB)=1+sinC and a=4,b=3, then area (in sq. units) of the triangle ABC is

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Solution

sinA(sinA+cosBsinB)+cosA(cosA+sinB+cosB)=1+sinC
(sin2A+cos2A)+(sinAcosB+cosAsinB)+(cosAcosBsinAsinB)=1+sinC
1+sin(A+B)+cos(A+B)=1+sinC
1+sinCcosC=1+sinC (A+B+C=π)
cosC=0
C=π2Δ=12absinC=12×4×3=6

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