Question Number 146048 by ArielVyny last updated on 10/Jul/21
$${in}\:{a}\:{triangle}\:{ABC}\:\:{we}\:{have}\: \\ $$$$\begin{cases}{\mathrm{3}{sin}\hat {{A}}+\mathrm{4}{cos}\hat {{B}}=\mathrm{6}}\\{\mathrm{4}{sin}\hat {{B}}+\mathrm{3}{cos}\hat {{A}}=\mathrm{1}}\end{cases} \\ $$$${find}\:\hat {{C}} \\ $$$$ \\ $$
Answered by gsk2684 last updated on 10/Jul/21
$${squaring}\:{and}\:{adding}\:{to}\:{get}\: \\ $$$$\mathrm{25}+\mathrm{24}\:\mathrm{sin}\:\left({A}+{B}\right)=\mathrm{37} \\ $$$$\mathrm{sin}\:\left(\mathrm{180}^{\mathrm{0}} −{C}\right)=\frac{\mathrm{12}}{\mathrm{24}} \\ $$$$\mathrm{sin}\:{C}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{C}=\frac{\Pi}{\mathrm{6}} \\ $$