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Question Number 64971 by Tawa1 last updated on 23/Jul/19

Commented by Tony Lin last updated on 23/Jul/19

$$\frac{\frac{x}{2}}{sin\theta }=\frac{\frac{x}{2}+1}{sin(90°+\theta )}=\frac{x}{sin(90°-2\theta )}$$$$\Rightarrow \frac{\frac{x}{2}}{sin\theta }=\frac{\frac{x}{2}+1}{cos\theta }=\frac{x}{cos2\theta }$$$$letcos\theta =t$$$$\Rightarrow \frac{\frac{x}{2}}{\sqrt{1-t^2}}=\frac{\frac{x}{2}+1}{t}=\frac{x}{2t^2-1}$$$$\frac{t}{\sqrt{1-t^2}}=\frac{\frac{x}{2}+1}{\frac{x}{2}}=1+\frac{2}{x}\to \left(1\right)$$$$\frac{t^2}{1-t^2}={(1+\frac{2}{x})}^2$$$$t^2={(1+\frac{2}{x})}^2-{(1+\frac{2}{x})}^2{t}^2$$$$[{(1+\frac{2}{x})}^2+1]t^2={(1+\frac{2}{x})}^2$$$$\Rightarrow t^2=\frac{{(1+\frac{2}{x})}^2}{{(1+\frac{2}{x})}^2+1}$$$$\frac{\frac{x}{2}}{\sqrt{1-t^2}}=\frac{x}{2t^2-1}\to \left(2\right)$$$$2\sqrt{1-t^2}=2t^2-1$$$$4(1-t^2)=4t^4-4t^2+1$$$$\Rightarrow t^2=\frac{\sqrt{3}}{2}$$$$from\left(1\right)\mathrm{\&}\left(2\right)$$$$\Rightarrow \frac{\sqrt{3}}{2}=\frac{{(1+\frac{2}{x})}^2}{{(1+\frac{2}{x})}^2+1}$$$$\sqrt{3}=(2-\sqrt{3}){(1+\frac{2}{x})}^2$$$${(1+\frac{2}{x})}^2=\frac{\sqrt{3}}{2-\sqrt{3}}$$$$\Rightarrow x=AD=\frac{2}{\sqrt{\frac{\sqrt{3}}{2-\sqrt{3}}}-1}\fallingdotseq 1.297$$$$AC=\sqrt{{AD}^2+1-2ADcos(90°-\theta )}$$$$=\sqrt{{AD}^2+1-2ADsin\theta }$$$$\fallingdotseq 1.316$$$$pleasecheckifthereareanymistakes$$

Commented by Tony Lin last updated on 23/Jul/19

Commented by Tawa1 last updated on 23/Jul/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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