Question Number 204729 by Amidip last updated on 26/Feb/24
Answered by A5T last updated on 26/Feb/24
$$\frac{{sin}\left(\mathrm{2}{x}\right)}{{AD}}=\frac{{sin}\left(\mathrm{140}−\mathrm{2}{x}\right)}{{AB}};\frac{{sin}\left(\mathrm{140}−\mathrm{3}{x}\right)}{{AD}}=\frac{{sin}\left({x}\right)}{{DC}={AB}} \\ $$$$\Rightarrow\frac{{sin}\left(\mathrm{140}−\mathrm{3}{x}\right)}{{sin}\left({x}\right)}=\frac{{sin}\left(\mathrm{2}{x}\right)}{{sin}\left(\mathrm{140}−\mathrm{2}{x}\right)}\Rightarrow{x}=\mathrm{35}° \\ $$