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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=(3/2) OA=R−1 AB=2 ∠ABC=60°=(π/3) cos ∠OBC=(3/(2R)) cos ∠ABO=((R^2 +2^2 −(R−1)^2 )/(2×2×R))=((3+2R)/(4R)) ∠ABO=(π/3)−∠OBC cos ∠ABO=(1/2)cos ∠OBC+((√3)/2)sin ∠OBC ⇒((3+2R)/(4R))=(1/2)×(3/(2R))+((√3)/2)×((√(4R^2 −9))/(2R)) ⇒2R=(√(3(4R^2 −9))) ⇒8R^2 =27 ⇒R=((3(√6))/4)≈1.837

$${OB}={R} \\ $$$${BC}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${OA}={R}−\mathrm{1} \\ $$$${AB}=\mathrm{2} \\ $$$$\angle{ABC}=\mathrm{60}°=\frac{\pi}{\mathrm{3}} \\ $$$$\mathrm{cos}\:\angle{OBC}=\frac{\mathrm{3}}{\mathrm{2}{R}} \\ $$$$\mathrm{cos}\:\angle{ABO}=\frac{{R}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} −\left({R}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}×{R}}=\frac{\mathrm{3}+\mathrm{2}{R}}{\mathrm{4}{R}} \\ $$$$\angle{ABO}=\frac{\pi}{\mathrm{3}}−\angle{OBC} \\ $$$$\mathrm{cos}\:\angle{ABO}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\angle{OBC}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\angle{OBC} \\ $$$$\Rightarrow\frac{\mathrm{3}+\mathrm{2}{R}}{\mathrm{4}{R}}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{2}{R}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\sqrt{\mathrm{4}{R}^{\mathrm{2}} −\mathrm{9}}}{\mathrm{2}{R}} \\ $$$$\Rightarrow\mathrm{2}{R}=\sqrt{\mathrm{3}\left(\mathrm{4}{R}^{\mathrm{2}} −\mathrm{9}\right)} \\ $$$$\Rightarrow\mathrm{8}{R}^{\mathrm{2}} =\mathrm{27} \\ $$$$\Rightarrow{R}=\frac{\mathrm{3}\sqrt{\mathrm{6}}}{\mathrm{4}}\approx\mathrm{1}.\mathrm{837} \\ $$

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