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Question Number 38015 by ajfour last updated on 20/Jun/18

Commented by ajfour last updated on 20/Jun/18

$$Find\theta intermsof\mathit{d},\mathit{R}ifcircle$$$$hasallthreecolouredareasequal.$$

Commented by MrW3 last updated on 21/Jun/18

$$let\alpha =\mathrm{\angle }TCQ$$$$\frac{1}{3}\times \pi R^2=\frac{R^2}{2}(\alpha -\sin \alpha )$$$$\Rightarrow \sin \alpha =\alpha -\frac{2\pi }{3}$$$$\Rightarrow \alpha =149.27°$$$$$$$$R\cos \left(\frac{\alpha }{2}\right)=d\sin \theta$$$$\Rightarrow \theta ={\sin }^{-1}\left[\frac{R}{d}\cos \frac{\alpha }{2}\right]={\sin }^{-1}(0.265\frac{R}{d})$$

Commented by ajfour last updated on 21/Jun/18

$$Quiteshort\mathrm{\&}verynice.Thanks.$$$$Pleasetrythe△one,Sir.$$

Commented by ajfour last updated on 21/Jun/18

$$WhatisdintermsofRifthe$$$$areaP\stackrel{⌢}{ST}isalsoequalto\frac{\pi R^2}{3}.$$

Commented by MrW3 last updated on 21/Jun/18

$$let\beta =\mathrm{\angle }QCR$$$$\beta =2[\theta +\frac{\pi }{2}-\frac{\alpha }{2}]=2\theta +\pi -\alpha$$$$A_{PQRP}=2\times \frac{\pi R^2}{3}$$$$2[\frac{1}{2}{d}^2\sin \theta \cos \theta +\frac{1}{2}{R}^2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}]+\frac{R^2\beta }{2}=\frac{2\pi R^2}{3}$$$${\left(\frac{d}{R}\right)}^2\sin 2\theta +\sin \alpha +\beta =\frac{4\pi }{3}$$$${\left(\frac{d}{R}\right)}^2\sin 2\theta +\sin \alpha +2\theta +\pi -\alpha =\frac{4\pi }{3}$$$$since\alpha -\sin \alpha =\frac{2\pi }{3}$$$$\Rightarrow {\left(\frac{d}{R}\right)}^2\sin 2\theta +2\theta +\pi -\frac{2\pi }{3}=\frac{4\pi }{3}$$$$\Rightarrow {\left(\frac{d}{R}\right)}^2\sin 2\theta +2\theta =\pi$$$$since\theta ={\sin }^{-1}\left(\frac{\cos \frac{\alpha }{2}}{\frac{d}{R}}\right)={\sin }^{-1}\left(\frac{\cos \frac{\alpha }{2}}{\delta }\right)$$$$let\delta =\frac{d}{R}$$$$\sin 2\theta =2\sin \theta \cos \theta =2\times \frac{\cos \frac{\alpha }{2}}{\delta }\times \sqrt{1-{\left(\frac{\cos \frac{\alpha }{2}}{\delta }\right)}^2}$$$$\Rightarrow {\delta }^{2}2\times \frac{\cos \frac{\alpha }{2}}{\delta }\times \sqrt{1-{\left(\frac{\cos \frac{\alpha }{2}}{\delta }\right)}^2}+{2\sin }^{-1}\left(\frac{\cos \frac{\alpha }{2}}{\delta }\right)=\pi$$$$\Rightarrow \delta \cos \frac{\alpha }{2}\times \sqrt{1-{\left(\frac{\cos \frac{\alpha }{2}}{\delta }\right)}^2}+{\sin }^{-1}\left(\frac{\cos \frac{\alpha }{2}}{\delta }\right)=\frac{\pi }{2}$$$$\Rightarrow \delta \cos \frac{\alpha }{2}\times \sqrt{1-{\left(\frac{\cos \frac{\alpha }{2}}{\delta }\right)}^2}=\frac{\pi }{2}-{\sin }^{-1}\left(\frac{\cos \frac{\alpha }{2}}{\delta }\right)$$$$\Rightarrow {\cos }^2\frac{\alpha }{2}\sqrt{1-{\left(\frac{\cos \frac{\alpha }{2}}{\delta }\right)}^2}=\frac{\cos \frac{\alpha }{2}}{\delta }(\frac{\pi }{2}-{\sin }^{-1}\left(\frac{\cos \frac{\alpha }{2}}{\delta }\right))$$$$with\alpha =149.27°$$$$\Rightarrow \delta =5.7612$$$$\Rightarrow d=5.7612R$$

Commented by ajfour last updated on 21/Jun/18

$$Thankyousir.Wonderful$$$$solution.$$

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