Question Number 47947 by mr W last updated on 17/Nov/18
Commented by mr W last updated on 17/Nov/18
$${see}\:{Q}\mathrm{44017} \\ $$$$ \\ $$$${with}\:\lambda=\frac{{h}}{{R}}=\mathrm{tan}\:\varphi \\ $$$${a}={R}\frac{\lambda}{\mathrm{sin}\:\theta+\lambda\:\mathrm{cos}\:\theta} \\ $$$${b}={R}\sqrt{\frac{\lambda−\mathrm{tan}\:\theta}{\lambda+\mathrm{tan}\:\theta}} \\ $$