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Question Number 47947 by mr W last updated on 17/Nov/18

Commented by mr W last updated on 17/Nov/18

see Q44017    with λ=(h/R)=tan ϕ  a=R(λ/(sin θ+λ cos θ))  b=R(√((λ−tan θ)/(λ+tan θ)))

$${see}\:{Q}\mathrm{44017} \\ $$$$ \\ $$$${with}\:\lambda=\frac{{h}}{{R}}=\mathrm{tan}\:\varphi \\ $$$${a}={R}\frac{\lambda}{\mathrm{sin}\:\theta+\lambda\:\mathrm{cos}\:\theta} \\ $$$${b}={R}\sqrt{\frac{\lambda−\mathrm{tan}\:\theta}{\lambda+\mathrm{tan}\:\theta}} \\ $$

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