Question Number 188759 by Rupesh123 last updated on 06/Mar/23
Answered by a.lgnaoui last updated on 06/Mar/23
$$\left.\mathrm{D}\right)\:\mathrm{22} \\ $$
Answered by a.lgnaoui last updated on 07/Mar/23
$$\bigtriangleup{OMF}\:\:\:\:\mathrm{sin}\:\alpha=\frac{{HM}}{{MF}}=\frac{\mathrm{2}{R}}{{BC}}=\frac{\mathrm{5}}{\mathrm{6}}:\:\mathrm{cos}\:\alpha=\frac{\sqrt{\mathrm{11}}}{\mathrm{6}} \\ $$$$\Rightarrow{HM}=\frac{\mathrm{5}{MF}}{\mathrm{6}}\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\varphi=\left(\frac{\pi}{\mathrm{2}}−\alpha\right)\:\:\:\:\mathrm{sin}\:\varphi=\frac{{MF}}{{OF}}=\mathrm{cos}\:\alpha=\frac{\sqrt{\mathrm{11}}}{\mathrm{6}} \\ $$$$\left[\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{OF}=\frac{{MF}}{\mathrm{cos}\:\alpha}\right. \\ $$$${alors}\:\:\:\:\:{EF}={R}+\frac{{MF}}{\mathrm{cos}\:\alpha} \\ $$$$\:\:\:\:\:{OF}^{\mathrm{2}} ={MF}^{\mathrm{2}} +\mathrm{100}=\frac{{MF}^{\mathrm{2}} }{\mathrm{cos}\:^{\mathrm{2}} \alpha} \\ $$$$\:\:\:\:\:\:\:\mathrm{100}={MF}^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \alpha\:\:\:\:\:\:\:{MF}=\mathrm{2}\sqrt{\mathrm{11}}\: \\ $$$$\:\:\:\:\:\:\:\:{EF}={R}+\frac{\mathrm{2}\sqrt{\mathrm{11}}}{\:\sqrt{\mathrm{11}}}×\mathrm{6}=\mathrm{22}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\: \\ $$
Commented by a.lgnaoui last updated on 07/Mar/23
Commented by Rupesh123 last updated on 07/Mar/23
Excellent!
Commented by Shlock last updated on 07/Mar/23
Excellent!
Answered by manxsol last updated on 07/Mar/23
$$ \\ $$
Commented by manxsol last updated on 07/Mar/23
