Question Number 219450 by hardmath last updated on 25/Apr/25
Commented by mr W last updated on 25/Apr/25
$${the}\:{geometry}\:{is}\:{not}\:{uniquely}\:{defined}. \\ $$$${you}\:{can}'{t}\:{determine}\:\angle{CDO}\:{with} \\ $$$${given}\:{conditions}. \\ $$
Commented by hardmath last updated on 25/Apr/25
$$ \\ $$Find the angle between the axis of the cone inscribed in a triangle whose plane angles are equal to a and the angle of the triangle..
Commented by mr W last updated on 26/Apr/25
Answered by mr W last updated on 26/Apr/25
$${let}\:{AD}={BD}={CD}=\mathrm{1} \\ $$$${AB}={BC}={CA}=\mathrm{2}×\mathrm{1}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}=\mathrm{2}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}} \\ $$$${OC}=\frac{\mathrm{2}}{\mathrm{3}}×{CM}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\mathrm{2}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\theta=\frac{{OC}}{{CD}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}} \\ $$$$\angle{CDO}=\theta=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}\right)\:\:\checkmark \\ $$
Commented by hardmath last updated on 26/Apr/25
$$\mathrm{cool}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{my}\:\mathrm{darling}\:\mathrm{professor} \\ $$