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Question Number 173948 by dragan91 last updated on 21/Jul/22

Commented by dragan91 last updated on 21/Jul/22

$$\mathrm{Value}\mathrm{of}\mathrm{Angle}?$$

Commented by a.lgnaoui last updated on 22/Jul/22

$$2\alpha +\beta =\frac{\pi }{2}andx+\alpha =\frac{\pi }{2}$$$$\cos 2\alpha =\frac{1}{2}2\alpha =\frac{\pi }{3}\alpha =\frac{\pi }{6}$$$$x=\frac{\pi }{2}-\frac{\pi }{6}x=\frac{\pi }{3}$$

Commented by Tawa11 last updated on 22/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by mr W last updated on 23/Jul/22

$$howareyousurethat\cos 2\alpha =\frac{1}{2}?$$$$2\alpha +\beta =\frac{\pi }{2}isnottrue!$$$$\left(wronglymarkedinthequestion\right)$$$$youhaveignoredthedottedline.$$

Answered by mr W last updated on 23/Jul/22

Commented by mr W last updated on 23/Jul/22

$$1^2+{(2-r)}^2={(1+r)}^2$$$$\Rightarrow r=\frac{2}{3}$$$$\tan \theta =\frac{2-r}{1}=\frac{4}{3}=\frac{2\tan \frac{\theta }{2}}{1-{\tan }^2\frac{\theta }{2}}$$$$2{\tan }^2\frac{\theta }{2}+3\tan \frac{\theta }{2}-2=0$$$$(2\tan \frac{\theta }{2}-1)(\tan \frac{\theta }{2}+2)=0$$$$\Rightarrow \tan \frac{\theta }{2}=\frac{1}{2}\Rightarrow \sin \frac{\theta }{2}=\frac{1}{\sqrt{5}}$$$$$$$$2\alpha +\theta =\pi$$$$\Rightarrow \alpha =\frac{\pi }{2}-\frac{\theta }{2}$$$$2\beta +(\frac{\pi }{2}-\theta )=\pi$$$$\Rightarrow \beta =\frac{\pi }{4}+\frac{\theta }{2}$$$$\gamma =\pi -\alpha -\beta =\pi -\frac{\pi }{2}+\frac{\theta }{2}-\frac{\pi }{4}-\frac{\theta }{2}=\frac{\pi }{4}$$$$\Rightarrow EF=DE=2-OD=2-2\times 1\times \cos \alpha$$$$=2-2\cos (\frac{\pi }{2}-\frac{\theta }{2})$$$$=2-2\sin \frac{\theta }{2}$$$$=2(1-\frac{1}{\sqrt{5}})=\frac{2(\sqrt{5}-1)}{\sqrt{5}}$$$$\tan x=\frac{OE}{EF}=\frac{2\times \sqrt{5}}{2(\sqrt{5}-1)}=\frac{5+\sqrt{5}}{4}$$$$\Rightarrow x={\tan }^{-1}\frac{5+\sqrt{5}}{4}\approx 61.07°$$

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