Question Number 49365 by behi83417@gmail.com last updated on 06/Dec/18
Commented by behi83417@gmail.com last updated on 06/Dec/18
$${CA}=\mathrm{6},{AD}=\sqrt{\mathrm{3}},{DE}=\sqrt{\mathrm{2}},{EF}=\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:\:\:\boldsymbol{\mathrm{FC}}=? \\ $$
Answered by MJS last updated on 06/Dec/18
$$\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{intersecting}\:\mathrm{circles}\:\mathrm{with}\:“\mathrm{unfriendly}'' \\ $$$$\mathrm{coordinates}… \\ $$$${FC}\approx\mathrm{4}.\mathrm{59588} \\ $$
Answered by mr W last updated on 06/Dec/18
$${FC}=\mathrm{6}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}}{\mathrm{6}}−\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}}{\mathrm{6}}−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{6}}\right) \\ $$$$=\mathrm{6}\:\mathrm{cos}\:\left(\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}}{\mathrm{6}}+\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}}{\mathrm{6}}+\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{6}}\right) \\ $$$$\approx\mathrm{4}.\mathrm{6} \\ $$
Commented by behi83417@gmail.com last updated on 06/Dec/18
$${thank}\:{you}\:{very}\:{much}\:{sir}\:{MJS}\:{and}\:{sir}\:{mr}\:{W}. \\ $$