# In a triangle ABC ,if 3sin A+ 4cos B =6, and 4sin B +3cos A=1 then the possible value of angle C

Solution: We have ,

$\\ \Rightarrow 3sinA+4cosB=6 ........(1)\\ \\\Rightarrow 4cosB+3cosA=1 ........(2)$

Squaring and adding (1) and (2)  we get

$\\ 9+16+24\times sin(A+B)=37\\ \\\Rightarrow 24sin(A+B)=12\\ \\\Rightarrow sin(A+B)=1/2\\ \\ sin(\pi -C)=1/2\\ \\ \Rightarrow C=\pi /6$

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