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Question-49365




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=(√3),DE=(√2),EF=1  ⇒      FC=?
$${CA}=\mathrm{6},{AD}=\sqrt{\mathrm{3}},{DE}=\sqrt{\mathrm{2}},{EF}=\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:\:\:\boldsymbol{\mathrm{FC}}=? \\ $$
Answered by MJS last updated on 06/Dec/18
it′s just intersecting circles with “unfriendly”  coordinates...  FC≈4.59588
$$\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{intersecting}\:\mathrm{circles}\:\mathrm{with}\:“\mathrm{unfriendly}'' \\ $$$$\mathrm{coordinates}… \\ $$$${FC}\approx\mathrm{4}.\mathrm{59588} \\ $$
Answered by mr W last updated on 06/Dec/18
FC=6 sin ((π/2)−sin^(−1) ((√3)/6)−sin^(−1) ((√2)/6)−sin^(−1) (1/6))  =6 cos (sin^(−1) ((√3)/6)+sin^(−1) ((√2)/6)+sin^(−1) (1/6))  ≈4.6
$${FC}=\mathrm{6}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}}{\mathrm{6}}−\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}}{\mathrm{6}}−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{6}}\right) \\ $$$$=\mathrm{6}\:\mathrm{cos}\:\left(\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}}{\mathrm{6}}+\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}}{\mathrm{6}}+\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{6}}\right) \\ $$$$\approx\mathrm{4}.\mathrm{6} \\ $$
Commented by behi83417@gmail.com last updated on 06/Dec/18
thank you very much sir MJS and sir mr W.
$${thank}\:{you}\:{very}\:{much}\:{sir}\:{MJS}\:{and}\:{sir}\:{mr}\:{W}. \\ $$

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