Category: Geometry

Question Number 199728 by Mingma last updated on 08/Nov/23 Answered by AST last updated on 08/Nov/23 $$\frac{{sin}\mathrm{120}°}{{BC}}=\frac{{sin}\mathrm{20}°}{{AB}}\Rightarrow{AB}=\frac{\mathrm{2}\sqrt{\mathrm{3}}{BCsin}\mathrm{20}°}{\mathrm{3}} \\ $$$$\frac{{sin}\mathrm{20}°}{{AB}}=\frac{{sin}\mathrm{40}°}{{AC}}\Rightarrow{AC}=\mathrm{2}{ABcos}\mathrm{20}°=\frac{\mathrm{2}\sqrt{\mathrm{3}}{BCsin}\mathrm{40}}{\mathrm{3}} \\ $$$$\frac{{sin}\left({x}\right)}{{DC}}=\frac{{sinADC}}{{AC}}=\frac{{sin}\left(\mathrm{140}−{x}\right)}{{AC}}\Rightarrow\frac{{AC}}{{DC}}=\frac{{sin}\left(\mathrm{140}−{x}\right)}{{sin}\left({x}\right)} \\ $$$$=\frac{{sin}\mathrm{40}}{{sin}\mathrm{20}}=\mathrm{2}{cos}\mathrm{20}=\frac{{sin}\mathrm{140}{cosx}−{sin}\left({x}\right){cos}\left(\mathrm{140}\right)}{{sinx}} \\ $$$$={sin}\mathrm{40}{tan}\left({x}\right)−{cos}\left(\mathrm{90}+\mathrm{50}\right)={sin}\mathrm{40}{tanx}+{sin}\mathrm{50}…

Question Number 199672 by Mingma last updated on 07/Nov/23 Answered by mr W last updated on 07/Nov/23 Commented by mr W last updated on 07/Nov/23…

Question Number 199510 by universe last updated on 04/Nov/23 Answered by mr W last updated on 06/Nov/23 Commented by mr W last updated on 06/Nov/23…

Question Number 199413 by universe last updated on 03/Nov/23 Answered by ajfour last updated on 03/Nov/23 $${b}\mathrm{cos}\:\alpha={R} \\ $$$$\left\{{R}\mathrm{tan}\:\alpha+\left(\frac{{a}}{{b}}\right){R}\right\}^{\mathrm{2}} +\left\{\left(\frac{{a}}{{b}}\right){R}\mathrm{tan}\:\alpha\right\}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${say}\:\:\:\mathrm{tan}\:\alpha={t} \\ $$$$\Rightarrow\:{t}^{\mathrm{2}}…

Question Number 199358 by ajfour last updated on 01/Nov/23 Commented by ajfour last updated on 01/Nov/23 $${Find}\:{R}. \\ $$ Answered by mr W last updated…