Question Number 193398 by Mingma last updated on 12/Jun/23
Answered by Subhi last updated on 13/Jun/23
$$ \\ $$$${put}\:\left({F}\hat {{C}E}\right)={y}=\left({C}\hat {{E}F}\right) \\ $$$$\left({F}\hat {{E}D}\right)=\left({F}\hat {{D}E}\right)=\left({E}\hat {{F}D}\right)=\mathrm{60} \\ $$$$\left({D}\hat {{E}B}\right)=\left({D}\hat {{B}E}\right)=\mathrm{180}−\mathrm{60}−{y}=\mathrm{120}−{y} \\ $$$$\left({B}\hat {{D}E}\right)=\left({E}\hat {{D}C}\right)\mathrm{180}−\mathrm{2}\left(\mathrm{120}−{y}\right)=\mathrm{180}−\mathrm{240}+\mathrm{2}{y}=\mathrm{2}{y}−\mathrm{60} \\ $$$$\left({A}\hat {{F}E}\right)={x}+\mathrm{60}=\mathrm{2}{y}\:\:\Rrightarrow\:{x}=\mathrm{2}{y}−\mathrm{60} \\ $$$$\left({F}\hat {{C}D}\right)=\left({F}\hat {{D}C}\right)=\frac{\mathrm{180}−\left(\mathrm{180}−\mathrm{2}{y}+\mathrm{60}\right)}{\mathrm{2}}={y}−\mathrm{30} \\ $$$$\left({F}\hat {{D}E}\right)=\left({F}\hat {{D}C}\right)+\left({C}\hat {{D}E}\right)=\mathrm{60}=\mathrm{2}{y}−\mathrm{60}+{y}−\mathrm{30} \\ $$$$\mathrm{3}{y}−\mathrm{90}=\mathrm{60}\:\:\Rrightarrow\:{y}=\mathrm{50} \\ $$$${x}=\mathrm{2}{y}−\mathrm{60}=\mathrm{2}×\mathrm{50}−\mathrm{60}\:=\:\mathrm{40} \\ $$
Commented by Mingma last updated on 13/Jun/23
Perfect ��