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Question Number 188759 by Rupesh123 last updated on 06/Mar/23

Answered by a.lgnaoui last updated on 06/Mar/23

$$\mathrm{D})22$$

Answered by a.lgnaoui last updated on 07/Mar/23

$$△OMF\sin \alpha =\frac{HM}{MF}=\frac{2R}{BC}=\frac{5}{6}:\cos \alpha =\frac{\sqrt{11}}{6}$$$$\Rightarrow HM=\frac{5MF}{6}\left(1\right)$$$$\phi =(\frac{\pi }{2}-\alpha )\sin \phi =\frac{MF}{OF}=\cos \alpha =\frac{\sqrt{11}}{6}$$$$[OF=\frac{MF}{\cos \alpha }$$$$alorsEF=R+\frac{MF}{\cos \alpha }$$$${OF}^2={MF}^2+100=\frac{{MF}^2}{\cos {}^2\alpha }$$$$100={MF}^2\tan {}^2\alpha MF=2\sqrt{11}$$$$EF=R+\frac{2\sqrt{11}}{\sqrt{11}}\times 6=22$$$$$$

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

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