Question-74474

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Question Number 74474 by ajfour last updated on 24/Nov/19

Commented by ajfour last updated on 24/Nov/19

$${Find}\:{the}\:{sides}\:{of}\:{the}\:{rectangle}. \\ $$

Answered by mr W last updated on 24/Nov/19

Commented by mr W last updated on 24/Nov/19

AD=cos α+(1/(sin α)) AB=sin α+(1/(cos α)) FB=(1/(cos α)) ((BC)/(FB))=tan ((π/4)+α) ((cos α+(1/(sin α)))/(1/(cos α)))=tan ((π/4)+α) ⇒(1/(1+tan^2 α))+(1/(tan α))=((1+tan α)/(1−tan α)) with t=tan α ⇒(1/(1+t^2 ))+(1/t)=((1+t)/(1−t)) ⇒((t^2 +t+1)/(t(1+t^2 )))=((1+t)/(1−t)) ⇒t^4 +2t^3 +t^2 +t−1=0 ⇒t=0.484 ⇒α=25.827°

$${AD}=\mathrm{cos}\:\alpha+\frac{\mathrm{1}}{\mathrm{sin}\:\alpha} \\ $$$${AB}=\mathrm{sin}\:\alpha+\frac{\mathrm{1}}{\mathrm{cos}\:\alpha} \\ $$$${FB}=\frac{\mathrm{1}}{\mathrm{cos}\:\alpha} \\ $$$$\frac{{BC}}{{FB}}=\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\alpha\right) \\ $$$$\frac{\mathrm{cos}\:\alpha+\frac{\mathrm{1}}{\mathrm{sin}\:\alpha}}{\frac{\mathrm{1}}{\mathrm{cos}\:\alpha}}=\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\alpha\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\alpha}+\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}=\frac{\mathrm{1}+\mathrm{tan}\:\alpha}{\mathrm{1}−\mathrm{tan}\:\alpha} \\ $$$${with}\:{t}=\mathrm{tan}\:\alpha \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{1}}{{t}}=\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}} \\ $$$$\Rightarrow\frac{{t}^{\mathrm{2}} +{t}+\mathrm{1}}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}=\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}} \\ $$$$\Rightarrow{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{3}} +{t}^{\mathrm{2}} +{t}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{0}.\mathrm{484}\:\Rightarrow\alpha=\mathrm{25}.\mathrm{827}° \\ $$

Commented by ajfour last updated on 25/Nov/19

$${Thank}\:{you}\:{Sir}. \\ $$

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