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Question Number 167008 by mnjuly1970 last updated on 04/Mar/22

Answered by mr W last updated on 06/Mar/22

Commented by Tawa11 last updated on 06/Mar/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 06/Mar/22

cos α=((30)/(2R))=((15)/R) β=π−2α sin β=sin 2α=((48)/(2R))=((24)/R) cos 2α=(√(1−(((24)/R))^2 ))=2(((15)/R))^2 −1 1−((24^2 )/R^2 )=4×((15^4 )/R^4 )−4×((15^2 )/R^2 )+1 R^2 =((15^4 )/(15^2 −12^2 )) ⇒R=25 (R−x)^2 +48^2 =(2R)^2 ⇒x=R−(√((2R)^2 −48^2 ))=25−14=11

$$\mathrm{cos}\:\alpha=\frac{\mathrm{30}}{\mathrm{2}{R}}=\frac{\mathrm{15}}{{R}} \\ $$$$\beta=\pi−\mathrm{2}\alpha \\ $$$$\mathrm{sin}\:\beta=\mathrm{sin}\:\mathrm{2}\alpha=\frac{\mathrm{48}}{\mathrm{2}{R}}=\frac{\mathrm{24}}{{R}} \\ $$$$\mathrm{cos}\:\mathrm{2}\alpha=\sqrt{\mathrm{1}−\left(\frac{\mathrm{24}}{{R}}\right)^{\mathrm{2}} }=\mathrm{2}\left(\frac{\mathrm{15}}{{R}}\right)^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{1}−\frac{\mathrm{24}^{\mathrm{2}} }{{R}^{\mathrm{2}} }=\mathrm{4}×\frac{\mathrm{15}^{\mathrm{4}} }{{R}^{\mathrm{4}} }−\mathrm{4}×\frac{\mathrm{15}^{\mathrm{2}} }{{R}^{\mathrm{2}} }+\mathrm{1} \\ $$$${R}^{\mathrm{2}} =\frac{\mathrm{15}^{\mathrm{4}} }{\mathrm{15}^{\mathrm{2}} −\mathrm{12}^{\mathrm{2}} } \\ $$$$\Rightarrow{R}=\mathrm{25} \\ $$$$\left({R}−{x}\right)^{\mathrm{2}} +\mathrm{48}^{\mathrm{2}} =\left(\mathrm{2}{R}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{x}={R}−\sqrt{\left(\mathrm{2}{R}\right)^{\mathrm{2}} −\mathrm{48}^{\mathrm{2}} }=\mathrm{25}−\mathrm{14}=\mathrm{11} \\ $$

Commented by mr W last updated on 06/Mar/22

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