Question Number 200475 by Rupesh123 last updated on 19/Nov/23
Answered by AST last updated on 19/Nov/23
$$\frac{\mathrm{2}{sin}\mathrm{20}}{{PC}}=\frac{\mathrm{1}}{{AC}}\Rightarrow\frac{{PC}}{{AC}}=\mathrm{2}{sin}\mathrm{20} \\ $$$${Let}\:\angle{PBC}={x};\:\frac{{sinx}}{{PC}}=\frac{{cos}\mathrm{10}}{{BC}}\Rightarrow{BC}=\frac{{PCcos}\left(\mathrm{10}°\right)}{{sin}\left({x}\right)} \\ $$$$\frac{{sin}\left(\mathrm{20}+{x}\right)}{{AC}}=\frac{{sin}\mathrm{50}}{{BC}}=\frac{{sin}\mathrm{50}{sinx}}{{PCcos}\mathrm{10}}\Rightarrow\frac{{PC}}{{AC}}=\frac{{sin}\mathrm{50}{sinx}}{{cos}\left(\mathrm{10}\right){sin}\left(\mathrm{20}+{x}\right)} \\ $$$$\Rightarrow\mathrm{2}{sin}\mathrm{20}=\frac{{sin}\mathrm{50}{sinx}}{{cos}\mathrm{10}{sin}\left(\mathrm{20}+{x}\right)}\Rightarrow\frac{{sinx}}{{sin}\left(\mathrm{20}+{x}\right)}=\frac{\mathrm{2}{sin}\mathrm{20}{cos}\mathrm{10}}{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \mathrm{20}} \\ $$$$\Rightarrow{x}=\mathrm{60}\Rightarrow\angle{PCB}=\mathrm{20}°\Rightarrow\angle{BCA}=\angle{BAC}=\mathrm{50}° \\ $$$$\Rightarrow{AB}={BC} \\ $$
Commented by Rupesh123 last updated on 19/Nov/23
Nice one, sir!