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

Commented by ajfour last updated on 24/Nov/19

$$Findthesidesoftherectangle.$$

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

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

$$AD=\cos \alpha +\frac{1}{\sin \alpha }$$$$AB=\sin \alpha +\frac{1}{\cos \alpha }$$$$FB=\frac{1}{\cos \alpha }$$$$\frac{BC}{FB}=\tan (\frac{\pi }{4}+\alpha )$$$$\frac{\cos \alpha +\frac{1}{\sin \alpha }}{\frac{1}{\cos \alpha }}=\tan (\frac{\pi }{4}+\alpha )$$$$\Rightarrow \frac{1}{1+{\tan }^2\alpha }+\frac{1}{\tan \alpha }=\frac{1+\tan \alpha }{1-\tan \alpha }$$$$witht=\tan \alpha$$$$\Rightarrow \frac{1}{1+t^2}+\frac{1}{t}=\frac{1+t}{1-t}$$$$\Rightarrow \frac{t^2+t+1}{t(1+t^2)}=\frac{1+t}{1-t}$$$$\Rightarrow t^4+2t^3+t^2+t-1=0$$$$\Rightarrow t=0.484\Rightarrow \alpha =25.827°$$

Commented by ajfour last updated on 25/Nov/19

$$ThankyouSir.$$

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