Question Number 171765 by cherokeesay last updated on 20/Jun/22
Answered by som(math1967) last updated on 21/Jun/22
$${let}\:{OB}={OA}={OM}={r} \\ $$$$\:\therefore{AC}=\sqrt{{r}^{\mathrm{2}} +\frac{{r}^{\mathrm{2}} }{\mathrm{4}}}=\frac{{r}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\bigtriangleup{AOC}\sim\bigtriangleup{OLC} \\ $$$$\:\therefore\frac{{LC}}{{OC}}=\frac{\frac{{r}}{\mathrm{2}}}{\frac{{r}\sqrt{\mathrm{5}}}{\mathrm{2}}} \\ $$$$\Rightarrow{LC}=\frac{{r}}{\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$$\therefore{AL}=\frac{{r}\sqrt{\mathrm{5}}}{\mathrm{2}}\:−\frac{{r}}{\mathrm{2}\sqrt{\mathrm{5}}}=\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{5}}} \\ $$$${same}\:{way} \\ $$$$\frac{{OL}}{{OC}}=\frac{{r}}{\frac{{r}\sqrt{\mathrm{5}}}{\mathrm{2}}} \\ $$$${OL}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}×\frac{{r}}{\mathrm{2}}=\frac{{r}}{\:\sqrt{\mathrm{5}}} \\ $$$${LM}={r}−\frac{{r}}{\:\sqrt{\mathrm{5}}}=\frac{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){r}}{\:\sqrt{\mathrm{5}}} \\ $$$$\:\therefore\:{tan}\beta=\frac{{AL}}{{LM}}=\frac{\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{5}}}}{\frac{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){r}}{\:\sqrt{\mathrm{5}}}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}−\mathrm{1}} \\ $$$$\therefore\:{tan}\beta=\frac{\mathrm{2}\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)}{\mathrm{5}−\mathrm{1}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}=\varphi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{golden}}\:\boldsymbol{{ratio}} \\ $$$$ \\ $$
Commented by som(math1967) last updated on 21/Jun/22
Commented by Tawa11 last updated on 21/Jun/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by cherokeesay last updated on 21/Jun/22
$${Nice}\:….{thank}\:{you}\:{sir}. \\ $$
Answered by mr W last updated on 21/Jun/22
Commented by mr W last updated on 21/Jun/22
$$\mathrm{tan}\:\alpha=\frac{\mathrm{2}}{\mathrm{1}}=\frac{\mathrm{2}\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}} \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\varphi \\ $$$$ \\ $$$$\alpha+\mathrm{2}\beta=\pi\:\Rightarrow\beta=\frac{\pi}{\mathrm{2}}−\frac{\alpha}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{tan}\:\beta=\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\varphi\:\checkmark \\ $$
Commented by Tawa11 last updated on 21/Jun/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by cherokeesay last updated on 21/Jun/22
$${thank}\:{you}\:{mister}. \\ $$