Question Number 166063 by cortano1 last updated on 12/Feb/22
Commented by blackmamba last updated on 12/Feb/22
$$\:\mathrm{tan}\:\alpha=\frac{\mathrm{2}}{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\mathrm{3}−\sqrt{\mathrm{3}}}=\frac{\mathrm{3}+\sqrt{\mathrm{3}}}{\mathrm{6}} \\ $$$$\:\mathrm{tan}\:\mathrm{2}\alpha=\frac{\mathrm{2}\left(\mathrm{15}+\mathrm{7}\sqrt{\mathrm{3}}\right)}{\mathrm{13}} \\ $$$$\:\mathrm{tan}\:\left(\mathrm{90}°−\mathrm{2}\alpha\right)=\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{2}\alpha}=\frac{\mathrm{15}−\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{12}} \\ $$$$\:{m}=\frac{\mathrm{15}−\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{12}} \\ $$
Commented by MaxiMaths last updated on 12/Feb/22
$$\mathrm{cool}... \\ $$