Question Number 222376 by ajfour last updated on 24/Jun/25
Answered by ajfour last updated on 24/Jun/25
$${Say}\:{AB}\:={s} \\ $$$${AF}={t} \\ $$$${height}\:{k} \\ $$$${horizontal}\:{projestion}\:{of}\:{CD}={h} \\ $$$$\mathrm{cos}\:\theta=\frac{{x}^{\mathrm{4}} +{c}^{\mathrm{2}} −{p}^{\mathrm{2}} }{\mathrm{2}{cx}^{\mathrm{2}} }=\frac{\mathrm{1}+{x}^{\mathrm{2}} −{s}^{\mathrm{2}} }{\mathrm{2}{x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\:\phi}=\frac{{s}}{\mathrm{sin}\:\theta}=\frac{{x}}{\mathrm{sin}\:\left(\theta+\phi\right)} \\ $$$$\left({s}+{t}\right)^{\mathrm{2}} ={h}^{\mathrm{2}} +{k}^{\mathrm{2}} \\ $$$$\mathrm{cos}\:\phi=\frac{\left({c}+{x}\right)^{\mathrm{2}} +\left({s}+{t}\right)^{\mathrm{2}} −{x}^{\mathrm{4}} }{\mathrm{2}\left({c}+{x}\right)\left({s}+{t}\right)}=\frac{{x}^{\mathrm{2}} +{s}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{sx}} \\ $$$$\left({s}+{t}\right)\mathrm{cos}\:\left(\theta+\phi\right)={k} \\ $$$$\left({s}+{t}\right)\mathrm{sin}\:\left(\theta+\phi\right)={h} \\ $$$$… \\ $$$$ \\ $$