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Question Number 65365 by aditya@345 last updated on 29/Jul/19

$$3\mathrm{sinA}+4\mathrm{cosB}=6$$$$3\mathrm{cosA}+4\mathrm{sinB}=1$$$$\mathrm{find}\mathrm{angle}\mathrm{C}$$

Answered by Tanmay chaudhury last updated on 29/Jul/19

$${(3sinA+4cosB)}^2+{(3cosA+4sinB)}^2=37$$$$9({sin}^{2}A+{cos}^{2}A)+16({cos}^{2}B+{sin}^{2}B)+24(sinAcosB+cosAsinB)=37$$$$24sin(A+B)=37-25$$$$sin(\pi -C)=\frac{12}{24}=\frac{1}{2}$$$$sinC=\frac{1}{2}=sin\frac{\pi }{6}C=\frac{\pi }{6}$$

Answered by behi83417@gmail.com last updated on 29/Jul/19

$${9\sin }^2\mathrm{A}+{16\cos }^2\mathrm{B}+24\mathrm{sinA}.\mathrm{cosB}=36$$$${9\cos }^2\mathrm{A}+{16\sin }^2\mathrm{B}+24\mathrm{cosAsinB}=1$$$$-+++------++++++---$$$$9+16+24\sin (\mathrm{A}+\mathrm{B})=37$$$$24\sin (180-\mathrm{C})=12$$$$\Rightarrow \mathrm{sinC}=\frac{1}{2}\Rightarrow \stackrel{}{\mathrm{C}}={30}^{\bullet }.◼$$

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