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Question Number 75518 by aliesam last updated on 12/Dec/19

Answered by mr W last updated on 12/Dec/19

Commented by mr W last updated on 12/Dec/19

$$OB=R$$$$BC=\frac{3}{2}$$$$OA=R-1$$$$AB=2$$$$\mathrm{\angle }ABC=60°=\frac{\pi }{3}$$$$\cos \mathrm{\angle }OBC=\frac{3}{2R}$$$$\cos \mathrm{\angle }ABO=\frac{R^2+2^2-{(R-1)}^2}{2\times 2\times R}=\frac{3+2R}{4R}$$$$\mathrm{\angle }ABO=\frac{\pi }{3}-\mathrm{\angle }OBC$$$$\cos \mathrm{\angle }ABO=\frac{1}{2}\cos \mathrm{\angle }OBC+\frac{\sqrt{3}}{2}\sin \mathrm{\angle }OBC$$$$\Rightarrow \frac{3+2R}{4R}=\frac{1}{2}\times \frac{3}{2R}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{4R^2-9}}{2R}$$$$\Rightarrow 2R=\sqrt{3(4R^2-9)}$$$$\Rightarrow 8R^2=27$$$$\Rightarrow R=\frac{3\sqrt{6}}{4}\approx 1.837$$

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