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Question Number 216332 by mr W last updated on 04/Feb/25

Answered by cherokeesay last updated on 04/Feb/25

Answered by A5T last updated on 04/Feb/25

Commented by A5T last updated on 04/Feb/25

$$\frac{1}{2}\times 6\times 10\times \sin \theta =\frac{6\times 10\times \sqrt{6^2+{10}^2-2\times 6\times 10\cos \theta }}{4\mathrm{R}}$$$$\Rightarrow 2\mathrm{Rsin}\theta =\sqrt{136-120\cos \theta }\Rightarrow \mathrm{R}=\frac{\sqrt{34-30\cos \theta }}{\sin \theta }$$$$\Rightarrow {\mathrm{R}}^2=\frac{34-30\cos \theta }{1-{\cos }^2\theta }...\left(\mathrm{i}\right)$$$$\frac{\sin (180-\theta )}{\mathrm{AB}}=\frac{\sin 90}{10}\Rightarrow \mathrm{AB}=10\sin \theta$$$${\left(2\mathrm{R}\right)}^2={\left(2\mathrm{AB}\right)}^2+6^2$$$$\Rightarrow {\mathrm{R}}^2={100\sin }^2\theta +9=109-{100\cos }^2\theta ...\left(\mathrm{ii}\right)$$$$\left(\mathrm{i}\right)\mathrm{\&}\left(\mathrm{ii}\right)\Rightarrow 109-{100\cos }^2\theta =\frac{34-30\cos \theta }{1-{\cos }^2\theta }$$$$\Rightarrow \cos \theta =\frac{3\underset{-}{+}\sqrt{209}}{20}$$$$\Rightarrow {\mathrm{R}}^2=109-\frac{109\underset{-}{+}3\sqrt{209}}{2}=\frac{109\stackrel{-}{+}3\sqrt{209}}{2}$$$$\Rightarrow \mathrm{R}=\sqrt{\frac{109+3\sqrt{209}}{2}}=\frac{3+\sqrt{209}}{2}\approx 8.728$$

Commented by mahdipoor last updated on 04/Feb/25

$$why({\left(2R\right)}^2={\left(2AB\right)}^2+6^2)?$$$$$$

Commented by A5T last updated on 04/Feb/25

$$\mathrm{The}\mathrm{diameter},\mathrm{length}2\mathrm{R},\mathrm{subtends}\mathrm{an}\mathrm{angle}\mathrm{of}$$$$90°\mathrm{at}\mathrm{the}\mathrm{circumference}.$$

Commented by A5T last updated on 04/Feb/25

Commented by A5T last updated on 04/Feb/25

$$\mathrm{CE}=2\mathrm{R};\mathrm{AE}=2\mathrm{AB};\mathrm{AC}=6$$$$\mathrm{\angle }\mathrm{CAE}=90°\Rightarrow {\mathrm{CE}}^2={\mathrm{AE}}^2={\mathrm{AC}}^2$$$$\Rightarrow {\left(2\mathrm{R}\right)}^2={\left(2\mathrm{AB}\right)}^2+6^2$$

Commented by mr W last updated on 04/Feb/25

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Answered by mr W last updated on 04/Feb/25

Commented by mr W last updated on 04/Feb/25

$$R=x+3$$$$y^2={(x+3)}^2-3^2={10}^2-x^2$$$$x^2+3x-50=0$$$$\Rightarrow x=\frac{-3+\sqrt{209}}{2}$$$$\Rightarrow R=\frac{-3+\sqrt{209}}{2}+3=\frac{3+\sqrt{209}}{2}✓$$

Answered by A5T last updated on 05/Feb/25

Commented by A5T last updated on 05/Feb/25

$$\mathrm{R}=3+10\cos (180-\theta )=3-10\cos \theta$$$$\mathrm{AB}=10\sin \theta =10\sqrt{1-{\cos }^2\theta }$$$${\left(2\mathrm{R}\right)}^2=6^2+{\left(2\mathrm{AB}\right)}^2\Rightarrow {\mathrm{R}}^2=3^2+{\mathrm{AB}}^2$$$$\Rightarrow {10\cos }^2\theta -3\cos \theta -5=0\Rightarrow \cos \theta =\frac{3\underset{-}{+}\sqrt{209}}{20}$$$$\Rightarrow \mathrm{R}=\frac{3+\sqrt{209}}{2}$$

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