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Question Number 39985 by ajfour last updated on 14/Jul/18

Answered by ajfour last updated on 15/Jul/18

$$eq.oflineOBC:y=x$$$$eq.ofAEC:y\tan \theta =x-1$$$$ForpointC:$$$$x_C=y_C=\frac{1}{1-\tan \theta }$$$$\mathrm{\angle }ADB=\theta$$$$soAD=\cot \theta$$$$slopeofCDbem$$$$-m=\frac{y_C}{1+\cot \theta -x_C}$$$$=\frac{1}{(1+\cot \theta )(1-\tan \theta )-1}$$$$=\frac{1}{-\tan \theta +\cot \theta -1}$$$$\tan \mathrm{\angle }DAF=\frac{1}{m}=1+\tan \theta -\cot \theta$$$$FD=AD\sin \mathrm{\angle }DAF$$$$=\cot \theta \times \frac{1+\tan \theta -\cot \theta }{\sqrt{1+{(1+\tan \theta -\cot \theta )}^2}}$$$$\mathrm{\angle }BCE=\frac{\pi }{4}-\theta$$$$CE=BE\cot (\frac{\pi }{4}-\theta )$$$$=\frac{\sin \theta }{\tan (\frac{\pi }{4}-\theta )}$$$$GivenCE=FD$$$$\Rightarrow \frac{\sin \theta }{\tan (\frac{\pi }{4}-\theta )}=\cot \theta \times \frac{1+\tan \theta -\cot \theta }{\sqrt{1+{(1+\tan \theta -\cot \theta )}^2}}$$$$\Rightarrow \frac{\sin \theta (1+\tan \theta )\tan \theta }{1-\tan \theta }=\frac{1+\tan \theta -\cot \theta }{\sqrt{1+{(1+\tan \theta -\cot \theta )}^2}}$$$$let\tan \theta =t\Rightarrow \sin \theta =\frac{t}{\sqrt{1+t^2}}$$$$\Rightarrow \frac{t(1+t)t}{\sqrt{1+t^2}(1-t)}=\frac{1+t-\frac{1}{t}}{\sqrt{1+{(1+t-\frac{1}{t})}^2}}$$$$\Rightarrow \frac{t^4{(1+t)}^2}{(1+t^2){(1-t)}^2}=\frac{{(t+t^2-1)}^2}{t^2+{(t+t^2-1)}^2}$$$$\Rightarrow t=\tan \theta \approx 0.63923$$$$\theta \approx 32.588°.$$

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