Question-224336

Question Number 224336 by wongb1506 last updated on 04/Sep/25

Answered by som(math1967) last updated on 04/Sep/25

let EC=x, ∠BEC=α ∴∠AED=180−60−α=120−α (4/x)=sinα⇒(1/x)=((sin α)/4) (3/x)=sin(120−α)⇒(1/x)=((sin(120−α))/3) ∴ ((sinα)/4)=((sin(120−α))/3) ⇒3sinα=4sin120cosα−4cos120sinα ⇒3sinα=2(√3)cosα+2sinα ⇒sinα=2(√3)cosα ∴tanα=2(√3)⇒sinα=((2(√3))/( (√(13)))) (1/x)=((2(√3))/(4(√(13)))) ⇒x=((2(√(13)))/( (√3))) area of △DEC=((√3)/4)×((52)/3)=((13(√3))/3)squnit

$${let}\:{EC}={x},\:\angle{BEC}=\alpha \\ $$$$\therefore\angle{AED}=\mathrm{180}−\mathrm{60}−\alpha=\mathrm{120}−\alpha \\ $$$$\:\frac{\mathrm{4}}{{x}}={sin}\alpha\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{\mathrm{sin}\:\alpha}{\mathrm{4}} \\ $$$$\:\:\frac{\mathrm{3}}{{x}}={sin}\left(\mathrm{120}−\alpha\right)\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{{sin}\left(\mathrm{120}−\alpha\right)}{\mathrm{3}} \\ $$$$\therefore\:\frac{{sin}\alpha}{\mathrm{4}}=\frac{{sin}\left(\mathrm{120}−\alpha\right)}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{3}{sin}\alpha=\mathrm{4}{sin}\mathrm{120}{cos}\alpha−\mathrm{4}{cos}\mathrm{120}{sin}\alpha \\ $$$$\Rightarrow\mathrm{3}{sin}\alpha=\mathrm{2}\sqrt{\mathrm{3}}{cos}\alpha+\mathrm{2}{sin}\alpha \\ $$$$\Rightarrow{sin}\alpha=\mathrm{2}\sqrt{\mathrm{3}}{cos}\alpha \\ $$$$\therefore{tan}\alpha=\mathrm{2}\sqrt{\mathrm{3}}\Rightarrow{sin}\alpha=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{13}}} \\ $$$$\:\:\frac{\mathrm{1}}{{x}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{4}\sqrt{\mathrm{13}}}\:\Rightarrow{x}=\frac{\mathrm{2}\sqrt{\mathrm{13}}}{\:\sqrt{\mathrm{3}}} \\ $$$$\:{area}\:{of}\:\bigtriangleup{DEC}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\frac{\mathrm{52}}{\mathrm{3}}=\frac{\mathrm{13}\sqrt{\mathrm{3}}}{\mathrm{3}}{squnit} \\ $$$$\:\: \\ $$$$ \\ $$

Answered by mr W last updated on 04/Sep/25

Commented by mr W last updated on 05/Sep/25

AF=(√(3^2 +4^2 −2×3×4 cos 60°))=(√(13)) DE=2R=((AF)/(sin 60°))=((2(√(13)))/( (√3))) [DEC]=(((√3)×(DE)^2 )/4)=((√3)/4)×((4×13)/3)=((13(√3))/( 3))

$${AF}=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} −\mathrm{2}×\mathrm{3}×\mathrm{4}\:\mathrm{cos}\:\mathrm{60}°}=\sqrt{\mathrm{13}} \\ $$$${DE}=\mathrm{2}{R}=\frac{{AF}}{\mathrm{sin}\:\mathrm{60}°}=\frac{\mathrm{2}\sqrt{\mathrm{13}}}{\:\sqrt{\mathrm{3}}} \\ $$$$\left[{DEC}\right]=\frac{\sqrt{\mathrm{3}}×\left({DE}\right)^{\mathrm{2}} }{\mathrm{4}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\frac{\mathrm{4}×\mathrm{13}}{\mathrm{3}}=\frac{\mathrm{13}\sqrt{\mathrm{3}}}{\:\mathrm{3}} \\ $$

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