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Question Number 117825 by snipers237 last updated on 13/Oct/20

Let ABC be a triangle such as    2cosA+3sinB=4 and  3cosB+2sinA=3  Prove that the angle C is right.

$${Let}\:{ABC}\:{be}\:{a}\:{triangle}\:{such}\:{as}\: \\ $$$$\:\mathrm{2}{cosA}+\mathrm{3}{sinB}=\mathrm{4}\:{and}\:\:\mathrm{3}{cosB}+\mathrm{2}{sinA}=\mathrm{3} \\ $$$${Prove}\:{that}\:{the}\:{angle}\:{C}\:{is}\:{right}. \\ $$$$\: \\ $$

Answered by john santu last updated on 13/Oct/20

2cos A+3sin B=4  2sin A+3cos B=3  ⇒4cos^2 A+12cos Asin B+9sin^2 B=16  ⇒4sin^2 A+12sin Acos B+9cos^2 B=9  (1)+(2)  ⇒13+12sin (A+B)=25  ⇒12sin (A+B)=12  ⇒sin (A+B)=1 ⇒A+B=90°    A+B+C=180°⇒90°+C=180°  then C = 90°

$$\mathrm{2cos}\:{A}+\mathrm{3sin}\:{B}=\mathrm{4} \\ $$$$\mathrm{2sin}\:{A}+\mathrm{3cos}\:{B}=\mathrm{3} \\ $$$$\Rightarrow\mathrm{4cos}\:^{\mathrm{2}} {A}+\mathrm{12cos}\:{A}\mathrm{sin}\:{B}+\mathrm{9sin}\:^{\mathrm{2}} {B}=\mathrm{16} \\ $$$$\Rightarrow\mathrm{4sin}\:^{\mathrm{2}} {A}+\mathrm{12sin}\:{A}\mathrm{cos}\:{B}+\mathrm{9cos}\:^{\mathrm{2}} {B}=\mathrm{9} \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{13}+\mathrm{12sin}\:\left({A}+{B}\right)=\mathrm{25} \\ $$$$\Rightarrow\mathrm{12sin}\:\left({A}+{B}\right)=\mathrm{12} \\ $$$$\Rightarrow\mathrm{sin}\:\left({A}+{B}\right)=\mathrm{1}\:\Rightarrow{A}+{B}=\mathrm{90}° \\ $$$$ \\ $$$${A}+{B}+{C}=\mathrm{180}°\Rightarrow\mathrm{90}°+{C}=\mathrm{180}° \\ $$$${then}\:{C}\:=\:\mathrm{90}° \\ $$

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