Solving problem with compound angle formula

$\displaystyle A+B+C={{180}^{\circ }}$

$\displaystyle 3\sin A+4\cos B=6$

$\displaystyle 3\cos A+4\sin B=\sqrt{{13}}$

$\displaystyle \sin x=?$

$\displaystyle \text{Solution}$

$\displaystyle A+B+C={{180}^{\circ }}$ 

$\displaystyle 3\sin A+4\cos B=6$ 

$\displaystyle {{\left( {3\sin A+4\cos B} \right)}^{2}}={{6}^{2}}$

$\displaystyle 9{{\sin }^{2}}A+24\sin A\cos B+16{{\cos }^{2}}B=36\ ..........(1)$

$\displaystyle 3\cos A+4\sin B=\sqrt{{13}}$

$\displaystyle {{\left( {3\cos A+4\sin B} \right)}^{2}}={{\left( {\sqrt{{13}}} \right)}^{2}}$ 

$\displaystyle 9{{\cos }^{2}}A+24\cos A\sin B+16{{\sin }^{2}}B=13\ ............(2)$ 

$\displaystyle (1)+(2)\Rightarrow \ $

$\displaystyle 9\left( {{{{\sin }}^{2}}A+{{{\cos }}^{2}}A} \right)+24\left( {\sin A\cos B+\cos A\sin B} \right)+16\left( {{{{\sin }}^{2}}B+{{{\cos }}^{2}}B} \right)=49$

$\displaystyle \ 9+24\sin \left( {A+B} \right)+16=49$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ 24\sin \left( {A+B} \right)=24$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sin \left( {A+B} \right)=1$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \sin \left( {{{{180}}^{\circ }}-C} \right)=1$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \ \ \sin C=1$