Question Number 192770 by ajfour last updated on 26/May/23
Commented by a.lgnaoui last updated on 26/May/23
$$\mathrm{question}\:\mathrm{not}\:\mathrm{clear}\: \\ $$
Commented by a.lgnaoui last updated on 26/May/23
$$\:\:\:\:\:\mathrm{graphe}\:\:\mathrm{must}\:\mathrm{be}\:\mathrm{explain}\:\mathrm{to}\:\mathrm{have}\:\mathrm{idea} \\ $$$$\mathrm{about}\:\mathrm{position}\:\mathrm{of}\:\boldsymbol{\mathrm{E}}\: \\ $$
Commented by ajfour last updated on 26/May/23
$${To}\:{the}\:{eye}\:{E},\:{find}\:{the}\:{ratio}\:{of}\:{the} \\ $$$${visible}\:{upper}\:{hemispherical}\:{area} \\ $$$${to}\:{the}\:{visible}\:{lower}\:{hemispherical} \\ $$$${area}.\:{E}\left({a},{b}\right)\:\:{if}\:{center}\:{be}\:{origin} \\ $$$${x}\:{axis}\:{along}\:{a}\:{and}\:{y}\:{axis}\:{along} \\ $$$${vertical}\:{axis}\:{of}\:{sphere}. \\ $$
Commented by mr W last updated on 18/Jun/23
$${question}\:{is}\:{very}\:{clear}! \\ $$
Answered by mr W last updated on 18/Jun/23
Commented by mr W last updated on 18/Jun/23
$${OF}={a} \\ $$$${FE}={b} \\ $$$${OE}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\mathrm{tan}\:\theta=\frac{{b}}{{a}} \\ $$$$\overset{\frown} {{ACB}}={total}\:{visible}\:{area} \\ $$$$\overset{\frown} {{ACD}}={visible}\:{area}\:{in}\:{upper}\:{hemisphere} \\ $$$$\overset{\frown} {{DB}}={visible}\:{area}\:{in}\:{lower}\:{hemisphere} \\ $$$$…… \\ $$