Question Number 65365 by aditya@345 last updated on 29/Jul/19
$$\mathrm{3sinA}+\mathrm{4cosB}=\mathrm{6} \\ $$$$\mathrm{3cosA}+\mathrm{4sinB}=\mathrm{1} \\ $$$$\mathrm{find}\:\mathrm{angle}\:\mathrm{C} \\ $$
Answered by Tanmay chaudhury last updated on 29/Jul/19
$$\left(\mathrm{3}{sinA}+\mathrm{4}{cosB}\right)^{\mathrm{2}} +\left(\mathrm{3}{cosA}+\mathrm{4}{sinB}\right)^{\mathrm{2}} =\mathrm{37} \\ $$$$\mathrm{9}\left({sin}^{\mathrm{2}} {A}+{cos}^{\mathrm{2}} {A}\right)+\mathrm{16}\left({cos}^{\mathrm{2}} {B}+{sin}^{\mathrm{2}} {B}\right)+\mathrm{24}\left({sinAcosB}+{cosAsinB}\right)=\mathrm{37} \\ $$$$\mathrm{24}{sin}\left({A}+{B}\right)=\mathrm{37}−\mathrm{25} \\ $$$${sin}\left(\pi−{C}\right)=\frac{\mathrm{12}}{\mathrm{24}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${sinC}=\frac{\mathrm{1}}{\mathrm{2}}={sin}\frac{\pi}{\mathrm{6}}\:\:{C}=\frac{\pi}{\mathrm{6}} \\ $$
Answered by behi83417@gmail.com last updated on 29/Jul/19
$$\mathrm{9sin}^{\mathrm{2}} \mathrm{A}+\mathrm{16cos}^{\mathrm{2}} \mathrm{B}+\mathrm{24sinA}.\mathrm{cosB}=\mathrm{36} \\ $$$$\mathrm{9cos}^{\mathrm{2}} \mathrm{A}+\mathrm{16sin}^{\mathrm{2}} \mathrm{B}+\mathrm{24cosAsinB}=\mathrm{1} \\ $$$$−+++−−−−−−++++++−−− \\ $$$$\mathrm{9}+\mathrm{16}+\mathrm{24sin}\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{37} \\ $$$$\mathrm{24sin}\left(\mathrm{180}−\mathrm{C}\right)=\mathrm{12} \\ $$$$\Rightarrow\mathrm{sinC}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\overset{ } {\mathrm{C}}=\mathrm{30}^{\bullet} \:\:\:\:\:.\blacksquare \\ $$