Category: Geometry

Question Number 200883 by Rupesh123 last updated on 25/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 200568 by Rupesh123 last updated on 20/Nov/23 Commented by AST last updated on 20/Nov/23 $$Cannotbeuniquelydetermined.$$ 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 $$\frac{2sin20}{PC}=\frac{1}{AC}\Rightarrow \frac{PC}{AC}=2sin20$$$$Let\mathrm{\angle }PBC=x;\frac{sinx}{PC}=\frac{cos10}{BC}\Rightarrow BC=\frac{PCcos\left(10°\right)}{sin\left(x\right)}$$$$\frac{sin(20+x)}{AC}=\frac{sin50}{BC}=\frac{sin50sinx}{PCcos10}\Rightarrow \frac{PC}{AC}=\frac{sin50sinx}{cos\left(10\right)sin(20+x)}$$$$\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 $$\cancel{\mathit{x}}$$ Answered by mr W last updated…