Question Number 199775 by Mingma last updated on 09/Nov/23
Answered by mathfreak01 last updated on 09/Nov/23
$$ \\ $$$$ \\ $$Assuming when have a triangle such as this $${DC}\:=\:\sqrt{{DB}^{\mathrm{2}} \:+\:{BC}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{5}} \\ $$$${AD}\:=\:{DC}\:=\:\sqrt{\mathrm{5}} \\ $$$$\therefore\:\angle{DAC}\:=\:\angle{ACD} \\ $$$$\angle{DAC}\:+\:\angle{ACD}\:=\:\angle{BDC}\:=\:\theta \\ $$$$\Rightarrow\:\angle{DAC}\:=\:\angle{ACD}\:=\:\frac{\theta}{\mathrm{2}} \\ $$$${From}\:\bigtriangleup{ABC}, \\ $$$$\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\:=\:\frac{\mathrm{2}}{\mathrm{1}\:+\:\sqrt{\mathrm{5}}} \\ $$$$\theta\:=\:\mathrm{2}\:\mathrm{arctan}\:\frac{\mathrm{2}}{\mathrm{1}\:+\:\sqrt{\mathrm{5}}} \\ $$$$ \\ $$$${From}\:\bigtriangleup{BDC}, \\ $$$$\mathrm{tan}\:\theta\:=\:\mathrm{2} \\ $$$$\theta\:=\:\mathrm{arctan}\:\mathrm{2} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{arctan}\:\mathrm{2}\:=\:\mathrm{2}\:\mathrm{arctan}\:\frac{\mathrm{2}}{\mathrm{1}\:+\:\sqrt{\mathrm{5}}} \\ $$$$ \\ $$
Commented by Mingma last updated on 09/Nov/23
Very elegant solution!
Answered by MM42 last updated on 09/Nov/23
$${tan}\left(\mathrm{2}{act}\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)\right)={tan}\left({act}\mathrm{2}\right) \\ $$$$\Rightarrow\frac{\mathrm{4}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{4}−\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{2} \\ $$$$\Rightarrow\mathrm{2}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)=\mathrm{4}−\mathrm{5}−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{5}}\:\:\checkmark \\ $$