Se da:
Paralelipipedul dreptunghic ABCDA'B'C'D' (Vezi desenul atasat.)
[tex]\bf~AB=6\sqrt{2}~cm\\AD=2\sqrt{2}~cm\\AA'=2\sqrt{6}~cm[/tex]
Se cere:
[tex]\bf\\m(\sphericalangle BC'(ABC))=?\\m(\sphericalangle AB'(ABC))=?[/tex]
Rezolvare:
Unghiul dintre o dreapta si un plan este egal cu unghiul dintre dreapta si proiectia ei pe acel plan.
Proiectia lui BC' pe planul ABC este muchia BC.
Proiectia lui AB' pe planul ABC este muchia AB.
[tex]\displaystyle\bf\\a)\\m(\sphericalangle BC'(ABC))=m(\sphericalangle C'BC)\\In~\Delta BCC'~avem:\\BC=AD=2\sqrt{2}~cm\\CC'=AA'=2\sqrt{6}~cm\\BC'=\sqrt{BC^2+CC'^2}=\sqrt{\Big(2\sqrt{2}\Big)^2+\Big(2\sqrt{6}\Big)^2}=\sqrt{8+24}=\sqrt{32}=4\sqrt{2}\\\\cos C'BC=\frac{BC}{BC'}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{2}{4}=\frac{1}{2}\\\\arccos\Big(\frac{1}{2}\Big)=60^o\\\\\implies~\boxed{\bf~m(\sphericalangle BC'(ABC))=m(\sphericalangle C'BC)=60^o}[/tex]
.
[tex]\displaystyle\bf\\b)\\m(\sphericalangle AB'(ABC))=m(\sphericalangle B'AB)\\In~\Delta ABB'~avem:\\AB=6\sqrt{2}~cm\\BB'=AA'=2\sqrt{6}~cm\\AB'=\sqrt{AB^2+BB'^2}=\sqrt{\Big(6\sqrt{2}\Big)^2+\Big(2\sqrt{6}\Big)^2}=\sqrt{72+24}=\sqrt{96}=4\sqrt{6}\\\\sin B'AC=\frac{BB'}{AB'}=\frac{2\sqrt{6}}{4\sqrt{6}}=\frac{2}{4}=\frac{1}{2}\\\\arcsin\Big(\frac{1}{2}\Big)=30^o\\\\\implies~\boxed{\bf~m(\sphericalangle AB'(ABC))=m(\sphericalangle B'AB)=30^o}[/tex]
