Question Number 128947 by ajfour last updated on 11/Jan/21
Commented by ajfour last updated on 11/Jan/21
$${Find}\:{radius}\:{of}\:{the}\:{blue}\:{arc}. \\ $$
Answered by mr W last updated on 11/Jan/21
Commented by mr W last updated on 11/Jan/21
$${a}={GF}+{FE}=\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${DE}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$$${AE}=\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$${AB}=\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:\mathrm{sin}\:\mathrm{30}°=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$${BC}=\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$${HB}=\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\mathrm{cos}\:\mathrm{30}°=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right) \\ $$$${HB}^{\mathrm{2}} =\left(\mathrm{2}{R}−{BC}\right){BC} \\ $$$$\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{4}}=\left(\mathrm{2}{R}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right)\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$\Rightarrow{R}=\frac{\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{4}}{\mathrm{2}}\approx\mathrm{4}.\mathrm{598076} \\ $$
Commented by ajfour last updated on 12/Jan/21
$${Thanks}\:{Sir}!\:\left({power}\:{of}\:{point}\right. \\ $$$$\left.{wrt}\:{a}\:{circle}\:{is}\:{really}\:{helpful}\right). \\ $$