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Question Number 32841 by alimudinade@gmail.com last updated on 03/Apr/18

$$A+B+C={180}^0$$$$3sinA+4cosB=6$$$$3cosA+4sinB=\sqrt{13}$$$$sinC=....$$

Answered by hknkrc46 last updated on 04/Apr/18

$$36+13=49$$$${(3\mathrm{sinA}+4\mathrm{cosB})}^2+{(3\mathrm{cosA}+4\mathrm{sinB})}^2=49$$$$9({\sin }^2\mathrm{A}+{\cos }^2\mathrm{A})+16({\sin }^2\mathrm{B}+{\cos }^2\mathrm{B})+24\sin (\mathrm{A}+\mathrm{B})=49$$$$9+16+24\sin (180-\mathrm{C})=49$$$$24\mathrm{sinC}=24$$$$\mathrm{sinC}=1$$$$$$

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