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Question-207946

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Question Number 207946 by cherokeesay last updated on 31/May/24
Answered by mr W last updated on 01/Jun/24
Commented by mr W last updated on 01/Jun/24
AC=(x/(cos θ)) BC=x tan θ CD=x(1−tan θ) ED=x(1−tan θ)tan θ CE=((x(1−tan θ))/(cos θ)) EG=x−x(1−tan θ)tan θ=x(1−tan θ+tan^2 θ) EF=(x/(cos θ))−((x(1−tan θ))/(cos θ))=((x tan θ)/(cos θ)) 5=(1/2)×(x/(cos θ))×((x(1−tan θ))/(cos θ)) ⇒10=x^2 (1−tan θ)(1+tan^2 θ) ...(i) 3=((sin ((π/2)−θ))/2)×x(1−tan θ+tan^2 θ)×((x tan θ)/(cos θ)) ⇒6=x^2 (1−tan θ+tan^2 θ)tan θ ...(ii) (i)/(ii): (5/3)=(((1−tan θ)(1+tan^2 θ))/((1−tan θ+tan^2 θ)tan θ)) let t=tan θ 8t^3 −8t^2 +8t−3=0 (2t−1)(4t^2 −2t+3)=0 ⇒2t−1=0 ⇒t=(1/2) 10=x^2 (1−(1/2))(1+(1/4)) ⇒x^2 =16 ⇒x=4 ✓
$${AC}=\frac{{x}}{\mathrm{cos}\:\theta} \\ $$$${BC}={x}\:\mathrm{tan}\:\theta \\ $$$${CD}={x}\left(\mathrm{1}−\mathrm{tan}\:\theta\right) \\ $$$${ED}={x}\left(\mathrm{1}−\mathrm{tan}\:\theta\right)\mathrm{tan}\:\theta \\ $$$${CE}=\frac{{x}\left(\mathrm{1}−\mathrm{tan}\:\theta\right)}{\mathrm{cos}\:\theta} \\ $$$${EG}={x}−{x}\left(\mathrm{1}−\mathrm{tan}\:\theta\right)\mathrm{tan}\:\theta={x}\left(\mathrm{1}−\mathrm{tan}\:\theta+\mathrm{tan}^{\mathrm{2}} \:\theta\right) \\ $$$${EF}=\frac{{x}}{\mathrm{cos}\:\theta}−\frac{{x}\left(\mathrm{1}−\mathrm{tan}\:\theta\right)}{\mathrm{cos}\:\theta}=\frac{{x}\:\mathrm{tan}\:\theta}{\mathrm{cos}\:\theta} \\ $$$$\mathrm{5}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{{x}}{\mathrm{cos}\:\theta}×\frac{{x}\left(\mathrm{1}−\mathrm{tan}\:\theta\right)}{\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\mathrm{10}={x}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{tan}\:\theta\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)\:\:\:…\left({i}\right) \\ $$$$\mathrm{3}=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right)}{\mathrm{2}}×{x}\left(\mathrm{1}−\mathrm{tan}\:\theta+\mathrm{tan}^{\mathrm{2}} \:\theta\right)×\frac{{x}\:\mathrm{tan}\:\theta}{\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\mathrm{6}={x}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{tan}\:\theta+\mathrm{tan}^{\mathrm{2}} \:\theta\right)\mathrm{tan}\:\theta\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)/\left({ii}\right): \\ $$$$\frac{\mathrm{5}}{\mathrm{3}}=\frac{\left(\mathrm{1}−\mathrm{tan}\:\theta\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)}{\left(\mathrm{1}−\mathrm{tan}\:\theta+\mathrm{tan}^{\mathrm{2}} \:\theta\right)\mathrm{tan}\:\theta} \\ $$$${let}\:{t}=\mathrm{tan}\:\theta \\ $$$$\mathrm{8}{t}^{\mathrm{3}} −\mathrm{8}{t}^{\mathrm{2}} +\mathrm{8}{t}−\mathrm{3}=\mathrm{0} \\ $$$$\left(\mathrm{2}{t}−\mathrm{1}\right)\left(\mathrm{4}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{t}−\mathrm{1}=\mathrm{0}\:\Rightarrow{t}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{10}={x}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\mathrm{16}\:\Rightarrow{x}=\mathrm{4}\:\checkmark \\ $$
Commented by mnjuly1970 last updated on 01/Jun/24
⋛
$$\:\underline{\underbrace{\lesseqgtr}} \\ $$
Commented by Tawa11 last updated on 21/Jun/24
Weldone sir
$$\mathrm{Weldone}\:\mathrm{sir} \\ $$

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