Question Number 222955 by ajfour last updated on 11/Jul/25
Commented by ajfour last updated on 11/Jul/25
$${Find}\:{the}\:{side}\:{length}\:{of}\:\:{square}\:{ABCD},\:{if} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{when}\:{it}\:{is}\:{a}\:{square}. \\ $$
Answered by mr W last updated on 12/Jul/25
$${GB}=\frac{\mathrm{1}}{\mathrm{cos}\:\theta} \\ $$$${AB}=\frac{\mathrm{1}}{\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta} \\ $$$${CB}=\left(\mathrm{cos}\:\theta+\frac{\mathrm{1}}{\mathrm{cos}\:\theta}\right)\frac{\mathrm{1}}{\mathrm{cos}\:\theta} \\ $$$$\left(\mathrm{cos}\:\theta+\frac{\mathrm{1}}{\mathrm{cos}\:\theta}\right)\frac{\mathrm{1}}{\mathrm{cos}\:\theta}=\frac{\mathrm{1}}{\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta} \\ $$$$\mathrm{cos}^{\mathrm{6}} \:\theta+\mathrm{cos}^{\mathrm{4}} \:\theta−\mathrm{1}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{6}} \:\theta}−\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\theta}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\theta}=\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{69}}}{\mathrm{18}}+\frac{\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{69}}}{\mathrm{18}}−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow\theta\approx\mathrm{29}.\mathrm{6762}° \\ $$$$\Rightarrow{CB}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}+\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{69}}}{\mathrm{18}}+\frac{\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{69}}}{\mathrm{18}}−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\approx\mathrm{2}.\mathrm{3247} \\ $$
Commented by mr W last updated on 12/Jul/25
$${thanks}! \\ $$
Commented by ajfour last updated on 12/Jul/25
$${yes},\:{all}\:{good}\:{now}! \\ $$