Question Number 207230 by mr W last updated on 10/May/24
Answered by mr W last updated on 10/May/24
Commented by mr W last updated on 10/May/24
$$\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{{y}}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}−{x}}{{y}}=\mathrm{60}° \\ $$$$\frac{\frac{{x}}{{y}}+\frac{\mathrm{4}−{x}}{{y}}}{\mathrm{1}−\frac{{x}\left(\mathrm{4}−{x}\right)}{{y}^{\mathrm{2}} }}=\sqrt{\mathrm{3}} \\ $$$$\frac{\mathrm{4}{y}}{{y}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{4}{x}}=\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{4}{x}=\frac{\mathrm{4}{y}}{\:\sqrt{\mathrm{3}}}−{y}^{\mathrm{2}} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{\mathrm{4}−{y}}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}−{x}}{\mathrm{4}−{y}}=\mathrm{105}° \\ $$$$\frac{\frac{{x}}{\mathrm{4}−{y}}+\frac{\mathrm{4}−{x}}{\mathrm{4}−{y}}}{\mathrm{1}−\frac{{x}\left(\mathrm{4}−{x}\right)}{\left(\mathrm{4}−{y}\right)^{\mathrm{2}} }}=−\sqrt{\mathrm{3}}−\mathrm{2} \\ $$$$\frac{\mathrm{4}\left(\mathrm{4}−{y}\right)}{{y}^{\mathrm{2}} −\mathrm{8}{y}+\mathrm{16}+{x}^{\mathrm{2}} −\mathrm{4}{x}}=−\sqrt{\mathrm{3}}−\mathrm{2} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{4}{x}=−\frac{\mathrm{4}\left(\mathrm{4}−{y}\right)}{\:\sqrt{\mathrm{3}}+\mathrm{2}}−{y}^{\mathrm{2}} +\mathrm{8}{y}−\mathrm{16} \\ $$$$\Rightarrow\frac{\mathrm{4}{y}}{\:\sqrt{\mathrm{3}}}−{y}^{\mathrm{2}} =−\frac{\mathrm{4}\left(\mathrm{4}−{y}\right)}{\:\sqrt{\mathrm{3}}+\mathrm{2}}−{y}^{\mathrm{2}} +\mathrm{8}{y}−\mathrm{16} \\ $$$$\Rightarrow{y}=\mathrm{3} \\ $$$$\frac{{green}}{{yellow}}=\frac{{y}}{\mathrm{4}−{y}}=\frac{\mathrm{3}}{\mathrm{4}−\mathrm{1}}=\mathrm{3}\:\checkmark \\ $$