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Question-172484

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Question Number 172484 by mnjuly1970 last updated on 27/Jun/22
Answered by mr W last updated on 28/Jun/22
let θ=∠ADC ((10)/(sin θ))=(5/(sin α))=((AD)/(sin (θ+α))) (6/(sin 2α))=((AD)/(sin (θ−2α))) sin θ=2 sin α AD=((5 sin (θ+α))/(sin α))=5(((sin θ)/(tan α))+cos θ) AD=((6 sin (θ−2α))/(sin 2α))=6(((sin θ)/(tan 2α))−cos θ) 5(((sin θ)/(tan α))+cos θ)=6(((sin θ)/(tan 2α))−cos θ) 11 cos θ=2 sin α((6/(tan 2α))−(5/(tan α))) 11 cos θ=−2((( 3−cos^2 α)/(cos α))) 121 cos^2 θ=4((( 9−6cos^2 α+cos^4 α)/(cos^2 α))) 121(1−4 sin^2 α)=4((( 9−6cos^2 α+cos^4 α)/(cos^2 α))) 121(4 cos^2 α−3)=4((( 9−6cos^2 α+cos^4 α)/(cos^2 α))) 40×4 cos^4 α−113 cos^2 α−12=0 cos^2 α=((113+(√(113^2 +4×40×4×12)))/(2×40×4))=(4/5) cos α=(2/( (√5))) ⇒sin α=(1/( (√5))) sin θ=2 sin α=(2/( (√5))) ⇒cos θ=−(1/( (√5))) sin ∠C=sin (π−α−θ)=sin (α+θ) =sin α cos θ+cos α sin θ =−(1/( (√5)))×(1/( (√5)))+(2/( (√5)))×(2/( (√5)))=(3/5) Area=((10×11)/2)×(3/5)=33
$${let}\:\theta=\angle{ADC} \\ $$$$\frac{\mathrm{10}}{\mathrm{sin}\:\theta}=\frac{\mathrm{5}}{\mathrm{sin}\:\alpha}=\frac{{AD}}{\mathrm{sin}\:\left(\theta+\alpha\right)} \\ $$$$\frac{\mathrm{6}}{\mathrm{sin}\:\mathrm{2}\alpha}=\frac{{AD}}{\mathrm{sin}\:\left(\theta−\mathrm{2}\alpha\right)} \\ $$$$\mathrm{sin}\:\theta=\mathrm{2}\:\mathrm{sin}\:\alpha \\ $$$${AD}=\frac{\mathrm{5}\:\mathrm{sin}\:\left(\theta+\alpha\right)}{\mathrm{sin}\:\alpha}=\mathrm{5}\left(\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\alpha}+\mathrm{cos}\:\theta\right) \\ $$$${AD}=\frac{\mathrm{6}\:\mathrm{sin}\:\left(\theta−\mathrm{2}\alpha\right)}{\mathrm{sin}\:\mathrm{2}\alpha}=\mathrm{6}\left(\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\mathrm{2}\alpha}−\mathrm{cos}\:\theta\right) \\ $$$$\mathrm{5}\left(\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\alpha}+\mathrm{cos}\:\theta\right)=\mathrm{6}\left(\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\mathrm{2}\alpha}−\mathrm{cos}\:\theta\right) \\ $$$$\mathrm{11}\:\mathrm{cos}\:\theta=\mathrm{2}\:\mathrm{sin}\:\alpha\left(\frac{\mathrm{6}}{\mathrm{tan}\:\mathrm{2}\alpha}−\frac{\mathrm{5}}{\mathrm{tan}\:\alpha}\right) \\ $$$$\mathrm{11}\:\mathrm{cos}\:\theta=−\mathrm{2}\left(\frac{\:\mathrm{3}−\mathrm{cos}^{\mathrm{2}} \:\alpha}{\mathrm{cos}\:\alpha}\right) \\ $$$$\mathrm{121}\:\mathrm{cos}^{\mathrm{2}} \:\theta=\mathrm{4}\left(\frac{\:\mathrm{9}−\mathrm{6cos}^{\mathrm{2}} \:\alpha+\mathrm{cos}^{\mathrm{4}} \:\alpha}{\mathrm{cos}^{\mathrm{2}} \:\alpha}\right) \\ $$$$\mathrm{121}\left(\mathrm{1}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\alpha\right)=\mathrm{4}\left(\frac{\:\mathrm{9}−\mathrm{6cos}^{\mathrm{2}} \:\alpha+\mathrm{cos}^{\mathrm{4}} \:\alpha}{\mathrm{cos}^{\mathrm{2}} \:\alpha}\right) \\ $$$$\mathrm{121}\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\alpha−\mathrm{3}\right)=\mathrm{4}\left(\frac{\:\mathrm{9}−\mathrm{6cos}^{\mathrm{2}} \:\alpha+\mathrm{cos}^{\mathrm{4}} \:\alpha}{\mathrm{cos}^{\mathrm{2}} \:\alpha}\right) \\ $$$$\mathrm{40}×\mathrm{4}\:\mathrm{cos}^{\mathrm{4}} \:\alpha−\mathrm{113}\:\mathrm{cos}^{\mathrm{2}} \:\alpha−\mathrm{12}=\mathrm{0} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\alpha=\frac{\mathrm{113}+\sqrt{\mathrm{113}^{\mathrm{2}} +\mathrm{4}×\mathrm{40}×\mathrm{4}×\mathrm{12}}}{\mathrm{2}×\mathrm{40}×\mathrm{4}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow\mathrm{sin}\:\alpha=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$\mathrm{sin}\:\theta=\mathrm{2}\:\mathrm{sin}\:\alpha=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow\mathrm{cos}\:\theta=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$\mathrm{sin}\:\angle{C}=\mathrm{sin}\:\left(\pi−\alpha−\theta\right)=\mathrm{sin}\:\left(\alpha+\theta\right) \\ $$$$=\mathrm{sin}\:\alpha\:\mathrm{cos}\:\theta+\mathrm{cos}\:\alpha\:\mathrm{sin}\:\theta \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${Area}=\frac{\mathrm{10}×\mathrm{11}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{5}}=\mathrm{33} \\ $$
Commented by Tawa11 last updated on 28/Jun/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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