Category: Geometry

Question Number 200568 by Rupesh123 last updated on 20/Nov/23 Commented by AST last updated on 20/Nov/23 $${Cannot}\:{be}\:{uniquely}\:{determined}. \\ $$ Commented by Frix last updated on…

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}} \\…

Question Number 200534 by obia last updated on 19/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 200457 by cortano12 last updated on 19/Nov/23 Commented by cortano12 last updated on 19/Nov/23 $$\:\:\:\:\cancel{\boldsymbol{{x}}} \\ $$ Answered by mr W last updated…

Question Number 200420 by ajfour last updated on 18/Nov/23 Commented by ajfour last updated on 18/Nov/23 $${plz}\:{solve}.. \\ $$ Commented by ajfour last updated on…