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Question Number 39089 by ajfour last updated on 02/Jul/18

Commented by ajfour last updated on 02/Jul/18

$$Find\frac{\mathit{a}}{\mathit{b}}and\mathit{\theta }.$$$$Thecurveisasemicircleand$$$$thetworedsegmentsareequal.$$

Answered by MJS last updated on 02/Jul/18

$$\mathrm{circle}:$$$${(x-\frac{a}{2})}^2+y^2={\left(\frac{a}{2}\right)}^2$$$$\mathrm{upper}\mathrm{semicircle}$$$$y=\sqrt{ax-x^2}$$$$\mathrm{tangent}\mathrm{in}P=\left(\begin{array}{c}p\\ \sqrt{ap-p^2}\end{array}\right)$$$$y=-\frac{2p-a}{2\sqrt{ap-p^2}}x+\frac{ap}{2\sqrt{ap-p^2}}$$$$(b=\frac{ap}{2\sqrt{ap-p^2}})$$$$Q=\left(\begin{array}{c}a\\ b\end{array}\right)$$$$\mathrm{line}PQ:$$$$y=\frac{p(2p-a)}{2(a-p)\sqrt{ap-p^2}}x+\frac{ap(2a-3p)}{2(a-p)\sqrt{ap-p^2}}$$$$y=0\Rightarrow x=\frac{ap(2a-3p)}{a-2p}$$$$x=a-p\Rightarrow$$$$\Rightarrow p=\frac{a\sqrt{2}}{2}$$$$\Rightarrow \mathrm{tangent}:$$$$y=-\frac{\sqrt{-2+2\sqrt{2}}}{2}x+\frac{\sqrt{1+\sqrt{2}}}{2}a$$$$\tan \theta =-\frac{\sqrt{-2+2\sqrt{2}}}{2}\Rightarrow \theta \approx -24.47°$$

Commented by ajfour last updated on 02/Jul/18

$$NoSir,\tan \theta =+\frac{\sqrt{-2+2\sqrt{2}}}{2}.$$

Commented by MJS last updated on 02/Jul/18

$$\mathrm{yes}\mathrm{of}\mathrm{course},\mathrm{if}\mathrm{you}\mathrm{measure}\theta \mathrm{as}\mathrm{a}\mathrm{positive}$$$$\mathrm{angle}.\mathrm{my}\mathrm{solution}\mathrm{comes}\mathrm{from}\mathrm{the}\mathrm{tangent}$$$$\mathrm{which}\mathrm{is}\mathrm{decreasing}...$$

Commented by ajfour last updated on 02/Jul/18

$$ThanksSir,ialsosolved(and$$$$ofcoursecreatedthequestion$$$$myself).$$

Commented by MJS last updated on 02/Jul/18

$$\mathrm{I}\mathrm{like}\mathrm{your}\mathrm{geometric}\mathrm{questions}\mathrm{and}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{very}$$$$\mathrm{interesting}\mathrm{how}\mathrm{different}\mathrm{people}\mathrm{differently}$$$$\mathrm{solve}\mathrm{the}\mathrm{same}\mathrm{problems}$$

Answered by ajfour last updated on 02/Jul/18

$$Letoriginbecenterofcircle.$$$$PointoftangencyP\equiv (\frac{a}{2}\sin \theta ,\frac{a}{2}\cos \theta )$$$$Thedashedlinemakesangle\theta$$$$withvertical;so$$$$\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{b-\frac{a}{2}\cos \theta }{\frac{a}{2}+\frac{a}{2}\sin \theta }$$$$\Rightarrow \frac{a}{2}\sin \theta +\frac{a}{2}\sin {}^2\theta =b\cos \theta -\frac{a}{2}\cos {}^2\theta$$$$b\cos \theta -\frac{a}{2}=\frac{a}{2}\sin \theta ...\left(i\right)$$$$lettheredsegmentbe$$$$x=\frac{a}{2}-\frac{a}{2}\sin \theta$$$$Then\frac{a-x}{b}=\frac{x}{b-y}$$$$\Rightarrow \frac{\frac{a}{2}+\frac{a}{2}\sin \theta }{b}=\frac{\frac{a}{2}-\frac{a}{2}\sin \theta }{b-\frac{a}{2}\cos \theta }$$$$b+b\sin \theta -\frac{a}{2}\cos \theta -\frac{a}{2}\cos \theta \sin \theta$$$$=b-b\sin \theta$$$$2b\sin \theta -\frac{a}{2}\cos \theta =\frac{a}{2}\sin \theta \cos \theta$$$$....\left(ii\right)$$$$Now\left(i\right)\times \cos \theta is$$$$b\cos {}^2\theta -\frac{a}{2}\cos \theta =\frac{a}{2}\sin \theta \cos \theta$$$$equatingthiswith\left(ii\right)$$$$2b\sin \theta =b\cos {}^2\theta$$$$\Rightarrow \sin {}^2\theta +2\sin \theta -1=0$$$${(\sin \theta +1)}^2=2$$$$\sin \theta =\pm \sqrt{2}-1$$$$consideringthe+vevalue$$$$\sin \theta =\sqrt{2}-1$$$$\tan \theta =\frac{\sqrt{2}-1}{\sqrt{1-{(\sqrt{2}-1)}^2}}$$$$\theta ={\tan }^{-1}\left(\frac{\sqrt{2}-1}{\sqrt{2\sqrt{2}-2}}\right)$$$$={\tan }^{-1}\left(\frac{\sqrt{2\sqrt{2}-2}}{2}\right)$$$$\frac{a}{b}=\frac{2\cos \theta }{1+\sin \theta }=\frac{2\sqrt{2\sqrt{2}-2}}{\sqrt{2}}$$$$hence\frac{\mathit{a}}{\mathit{b}}=2\sqrt{\sqrt{2}-1}.$$

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