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Question Number 97557 by Power last updated on 08/Jun/20

Answered by mr W last updated on 08/Jun/20

$$\mathrm{\angle }CBE=\mathrm{\angle }EBD=\theta =\frac{1}{2}(\frac{\pi }{2}-\alpha )$$$$\Rightarrow 2\theta =\frac{\pi }{2}-\alpha$$$$BD=2r\sin \alpha$$$$EB=\frac{2r}{\cos \theta }$$$$ED=\sqrt{{EB}^2+{BD}^2-2\times EB\times BD\times \cos \theta }$$$$ED=\sqrt{\frac{4r^2}{{\cos }^2\theta }+4r^2{\sin }^2\alpha -8r^2\sin \alpha }$$$$ED=2r\sqrt{\frac{1}{{\cos }^2\theta }+{\sin }^2\alpha -2\sin \alpha }$$$$\frac{\sin \phi }{EB}=\frac{\sin \theta }{ED}$$$$\sin \phi =\frac{EB\sin \theta }{ED}=\frac{2r\sin \theta }{\cos \theta 2r\sqrt{\frac{1}{{\cos }^2\theta }+{\sin }^2\alpha -2\sin \alpha }}$$$$\sin \phi =\frac{\sin \theta }{\sqrt{1+({\sin }^2\alpha -2\sin \alpha ){\cos }^2\theta }}$$$$\sin \phi =\frac{\sqrt{1-\cos 2\theta }}{\sqrt{2+({\sin }^2\alpha -2\sin \alpha )(\cos 2\theta +1)}}$$$$\sin \phi =\frac{\sqrt{1-\sin \alpha }}{\sqrt{2-(2-\sin \alpha )(1+\sin \alpha )\sin \alpha }}$$$$\Rightarrow \phi ={\sin }^{-1}\frac{\sqrt{1-\sin \alpha }}{\sqrt{2-(2-\sin \alpha )(1+\sin \alpha )\sin \alpha }}$$

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