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Question Number 49365 by behi83417@gmail.com last updated on 06/Dec/18

Commented by behi83417@gmail.com last updated on 06/Dec/18

$$CA=6,AD=\sqrt{3},DE=\sqrt{2},EF=1$$$$\Rightarrow \mathbf{FC}=?$$

Answered by MJS last updated on 06/Dec/18

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{just}\mathrm{intersecting}\mathrm{circles}\mathrm{with}‘‘{\mathrm{unfriendly}}^{″}$$$$\mathrm{coordinates}...$$$$FC\approx 4.59588$$

Answered by mr W last updated on 06/Dec/18

$$FC=6\sin (\frac{\pi }{2}-{\sin }^{-1}\frac{\sqrt{3}}{6}-{\sin }^{-1}\frac{\sqrt{2}}{6}-{\sin }^{-1}\frac{1}{6})$$$$=6\cos ({\sin }^{-1}\frac{\sqrt{3}}{6}+{\sin }^{-1}\frac{\sqrt{2}}{6}+{\sin }^{-1}\frac{1}{6})$$$$\approx 4.6$$

Commented by behi83417@gmail.com last updated on 06/Dec/18

$$thankyouverymuchsirMJSandsirmrW.$$

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