Category: Geometry

Question Number 200204 by cherokeesay last updated on 15/Nov/23 Commented by MM42 last updated on 15/Nov/23 $${is}\:,\:{ABCD},\:{is}\:{square}? \\ $$ Commented by ajfour last updated on…

Question Number 200092 by ajfour last updated on 13/Nov/23 Commented by ajfour last updated on 13/Nov/23 $${If}\:{semicircle}\:{has}\:{radius}\:\mathrm{1},\:{find} \\ $$$${radii}\:{of}\:{blue}\:{and}\:{green}\:{circles}. \\ $$$${One}\:{end}\:{of}\:{each}\:{semicircle}\:{is}\:{at} \\ $$$${center}\:{of}\:{other}. \\ $$…

Question Number 199911 by mr W last updated on 11/Nov/23 Commented by mr W last updated on 11/Nov/23 $${three}\:{mirrors}\:{build}\:{the}\:{sides}\:{of}\:{an} \\ $$$${equilateral}\:{triangle}.\:{a}\:{laser}\:{ray} \\ $$$${emitted}\:{from}\:{the}\:{midpoint}\:{of}\:{a}\:{side} \\ $$$${should}\:{reach}\:{the}\:{opposite}\:{vertex}\:{after}…

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…