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Question Number 81467 by ajfour last updated on 13/Feb/20

Commented by ajfour last updated on 13/Feb/20

$$Ifthecircularareacutoutis$$$$halfthesectorarea,find\alpha .$$

Answered by ajfour last updated on 13/Feb/20

$$(R-r)\sin \frac{\alpha }{2}=r$$$$\frac{R^2\alpha }{2}=2\pi r^2$$$$\Rightarrow {\left(\frac{R}{r}\right)}^2={\left[\frac{1+\sin \left(\alpha /2\right)}{\sin \left(\alpha /2\right)}\right]}^2=\frac{4\pi }{\alpha }$$$$or\sqrt{\alpha }(1+\mathrm{cosec}\frac{\alpha }{2})=2\sqrt{\pi }$$$$\Rightarrow \alpha =180°,29.231733°.$$

Commented by john santu last updated on 13/Feb/20

$$\alpha >{180}^{\mathrm{o}}??$$

Commented by ajfour last updated on 13/Feb/20

$$pleaselookcarefully,Sir.$$$$therearetwovalues,onis$$$$\alpha =180°.$$

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