Category: Geometry

Question Number 180547 by mnjuly1970 last updated on 13/Nov/22 Answered by mr W last updated on 13/Nov/22 Commented by mr W last updated on 13/Nov/22…

Question Number 49433 by behi83417@gmail.com last updated on 06/Dec/18 Commented by ajfour last updated on 06/Dec/18 $$circleradius,Sir?$$ Commented by behi83417@gmail.com last updated on…

Question Number 180459 by mnjuly1970 last updated on 12/Nov/22 Answered by greougoury555 last updated on 12/Nov/22 $$\cos 3\alpha =\frac{1}{2\cos \alpha }$$$$4\cos {}^3\alpha -3\cos \alpha =\frac{1}{2\cos \alpha }$$$$\cos \alpha =\frac{\sqrt{6+2\sqrt{17}}}{4}$$$$\:\mathrm{sin}\:\alpha\:=\:\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{17}}}}{\mathrm{4}} \…

Question Number 180458 by mnjuly1970 last updated on 12/Nov/22 Answered by mr W last updated on 13/Nov/22 $$c^2=1^2+{\left(\sqrt{3}\right)}^2+2\times 1\times \sqrt{3}\times \frac{1}{2}=4+\sqrt{3}$$$$\frac{\sin \alpha }{\sqrt{3}}=\frac{\sin 120°}{\sqrt{4+\sqrt{3}}}$$$$\mathrm{sin}\:\alpha=\sqrt{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}+\sqrt{\mathrm{3}}}}=\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{4}+\sqrt{\mathrm{3}}}}…

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…