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Question Number 64951 by naka3546 last updated on 23/Jul/19

Answered by mr W last updated on 23/Jul/19

Commented by mr W last updated on 23/Jul/19

$${\mathrm{\Delta }}_{BFC}=\frac{2\tan \beta }{2}=\tan \beta$$$$\frac{BD}{\sin \beta }=\frac{BC}{\sin 3\beta }$$$$\Rightarrow BD=\frac{2\sin \beta }{\sin 3\beta }$$$${\mathrm{\Delta }}_{BDC}=\frac{BC\times BD\sin 2\beta }{2}=\frac{2\sin \beta \sin 2\beta }{\sin 3\beta }$$$${\mathrm{\Delta }}_{BAC}=\frac{2\tan 2\beta }{2}=\tan 2\beta$$$$\mathrm{\Delta }ADFE={\mathrm{\Delta }}_{BAC}-{\mathrm{\Delta }}_{BFC}-2({\mathrm{\Delta }}_{BDC}-{\mathrm{\Delta }}_{BFC})$$$$={\mathrm{\Delta }}_{BAC}-2{\mathrm{\Delta }}_{BDC}+{\mathrm{\Delta }}_{BFC}=2{\mathrm{\Delta }}_{BFC}$$$$\Rightarrow {\mathrm{\Delta }}_{BAC}-2{\mathrm{\Delta }}_{BDC}={\mathrm{\Delta }}_{BFC}$$$$\Rightarrow \tan 2\beta -\frac{4\sin \beta \sin 2\beta }{\sin 3\beta }=\tan \beta$$$$\Rightarrow \frac{\sin 2\beta }{\cos 2\beta }-\frac{4\sin 2\beta }{3-4{\sin }^2\beta }=\frac{\sin \beta }{\cos \beta }$$$$\Rightarrow \frac{2}{2{\cos }^2\beta -1}-\frac{8}{4{\cos }^2\beta -1}=\frac{1}{{\cos }^2\beta }$$$$let\lambda ={\cos }^2\beta$$$$\Rightarrow \frac{2}{2\lambda -1}-\frac{8}{4\lambda -1}=\frac{1}{\lambda }$$$$\Rightarrow \frac{2(4\lambda -1)-8(2\lambda -1)}{(2\lambda -1)(4\lambda -1)}=\frac{1}{\lambda }$$$$\Rightarrow \frac{-8\lambda +6}{(2\lambda -1)(4\lambda -1)}=\frac{1}{\lambda }$$$$16{\lambda }^2-12\lambda +1=0$$$$\Rightarrow \lambda ={\cos }^2\beta =\frac{3\pm \sqrt{5}}{8}$$$$\Rightarrow \beta ={\cos }^{-1}\frac{\sqrt{6\pm 2\sqrt{5}}}{4}={\cos }^{-1}\frac{\sqrt{5}\pm 1}{4}$$$$\Rightarrow \beta =36°or72°$$$$\beta =36°issuitable$$$$\Rightarrow \alpha =180-2\times 2\times 36=36°$$

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