Question Number 84259 by M±th+et£s last updated on 10/Mar/20
Commented by M±th+et£s last updated on 10/Mar/20
$${find}\:\frac{{y}}{{x}} \\ $$$${and}\:{z} \\ $$
Answered by mr W last updated on 10/Mar/20
Commented by mr W last updated on 10/Mar/20
$${BD}={x}=\mathrm{2}{R}\:\mathrm{cos}\:\alpha \\ $$$${AC}={x}=\mathrm{2}{R}\:\mathrm{tan}\:\alpha \\ $$$$\Rightarrow\mathrm{cos}\:\alpha=\mathrm{tan}\:\alpha=\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\alpha} \\ $$$$\Rightarrow\mathrm{cos}^{\mathrm{2}} \:\alpha=\mathrm{sin}\:\alpha \\ $$$$\Rightarrow\mathrm{sin}^{\mathrm{2}} \:\alpha+\mathrm{sin}\:\alpha−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\alpha=\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\approx\mathrm{38}.\mathrm{17}° \\ $$$$\Rightarrow{z}=\gamma=\frac{\pi}{\mathrm{2}}−\alpha=\frac{\pi}{\mathrm{2}}−\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{z}=\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\approx\mathrm{51}.\mathrm{83}° \\ $$$$\Rightarrow\frac{{y}}{{x}}=\mathrm{cos}\:{z}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$
Commented by M±th+et£s last updated on 10/Mar/20
$${god}\:{bless}\:{you}\:{sir} \\ $$