Question Number 206804 by mr W last updated on 25/Apr/24
Answered by mr W last updated on 26/Apr/24
Commented by mr W last updated on 26/Apr/24
$$\frac{\mathrm{3}{y}}{\mathrm{5}{y}}×\frac{{a}}{{b}}×\frac{{c}+{d}}{{d}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{c}}{{d}}=\frac{\mathrm{5}{b}−\mathrm{3}{a}}{\mathrm{3}{a}}\:\:\:...\left({i}\right) \\ $$$$\frac{{x}}{\mathrm{3}{x}}×\frac{{d}}{{c}}×\frac{{b}+{a}}{{a}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{c}}{{d}}=\frac{{a}+{b}}{\mathrm{3}{a}}\:\:\:...\left({ii}\right) \\ $$$$\Rightarrow{a}+{b}=\mathrm{5}{b}−\mathrm{3}{a} \\ $$$$\Rightarrow{a}={b} \\ $$$$\frac{{BA}}{{BC}}=\frac{{b}}{{a}}=\mathrm{1}\:\Rightarrow{BA}={BC} \\ $$$$\Rightarrow\Delta{ABD}\equiv\Delta{CBD} \\ $$$$\Rightarrow\angle{A}=\angle{C}=\frac{\mathrm{180}−\mathrm{2}×\mathrm{35}}{\mathrm{2}}=\mathrm{55}°\:\checkmark \\ $$
Answered by A5T last updated on 26/Apr/24
Commented by mr W last updated on 26/Apr/24
��
Commented by A5T last updated on 26/Apr/24
$$\frac{{sin}\theta}{\mathrm{3}{x}}=\frac{{sin}\mathrm{35}}{\mathrm{3}{y}}\Rightarrow{sin}\theta=\frac{{xsin}\mathrm{35}}{{y}} \\ $$$$\frac{{sin}\theta}{{AB}}=\frac{{sin}\mathrm{70}}{\mathrm{8}{y}}\Rightarrow{AB}=\frac{\mathrm{4}{x}}{{cos}\mathrm{35}} \\ $$$${BC}^{\mathrm{2}} =\mathrm{16}{x}^{\mathrm{2}} +\frac{\mathrm{16}{x}^{\mathrm{2}} }{{cos}^{\mathrm{2}} \mathrm{35}}−\mathrm{32}{x}^{\mathrm{2}} =\frac{\mathrm{16}{x}^{\mathrm{2}} }{{cos}^{\mathrm{2}} \mathrm{35}}−\mathrm{16}{x}^{\mathrm{2}} \\ $$$$\Rightarrow{BC}=\mathrm{4}{x}\sqrt{\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \mathrm{35}}−\mathrm{1}}=\frac{\mathrm{4}{xsin}\mathrm{35}}{{cos}\mathrm{35}} \\ $$$$\frac{{sin}\left(\mathrm{35}+?\right)}{{AB}}=\frac{{sin}\mathrm{35}}{{BC}}\Rightarrow\frac{{sin}\left(\mathrm{35}+?\right){cos}\mathrm{35}}{\mathrm{4}{x}}=\frac{{cos}\mathrm{35}{sin}\mathrm{35}}{\mathrm{4}{xsin}\mathrm{35}} \\ $$$${sin}\left(\mathrm{35}+?\right)=\mathrm{1}\Rightarrow?=\mathrm{55}° \\ $$