Pe planul triunghiului ABC dreptunghic in A ridicam perpendicularele din punctele A si B , de aceeasi parte a planului , pe care consideram punctele M si N astfel incat BN<AM. Stiind ca \( AC=2a , AB=a\sqrt{3}, AM=a \) si ca planul MNC face cu planul ABC un unghi de \( 30^{\circ} \) , sa se afle:
a) aria triunghiului MNC;
b) distanta de la punctul B la planul MNC;
Drepte perpendiculare pe un plan
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Marius Mainea
- Gauss
- Posts: 1077
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- Location: Gaesti (Dambovita)
a)\( A_{\Delta ABC}=A_{\Delta MNC} \cdot cos 30 \Rightarrow A_{\Delta MNC}=2a^2 \)
b)Fie \( MN\cap AB=\{P\} \) si \( AT\perp PC \Rightarrow \) prin teorema celor trei perpendiculare \( MT\perp PC \);
\( MT\subset (MPC) \) si \( AT\subset(ABC) \Rightarrow \angle MTA=30 \) ; \( MT=2a \Rightarrow AT=a\sqrt{3} \)
In \( \Delta ACT \) ; \( sin \angle ACT =\frac{sqrt{3}}{2} \Rightarrow \angle ACT=60 \Rightarrow \) In \( \Delta APC \) ; \( tg60=\frac{AP}{2a} \Rightarrow AP=2a\sqrt{3} \Rightarrow PB=BA \Rightarrow BN=\frac{a}{2} \)
\( A_{\Delta MNB}=\frac{a^2\sqrt{3}}{4} \)
\( CA\perp AM \) si \( CA\perp AB \Rightarrow CA \perp (MNB) \)
\( V_{CMNB}=\frac{a^3\sqrt{3}}{6}=\frac{1}{3}\cdot A_{\Delta MNC} \cdot d \Rightarrow d=\frac{a\sqrt{3}}{4} \) .
b)Fie \( MN\cap AB=\{P\} \) si \( AT\perp PC \Rightarrow \) prin teorema celor trei perpendiculare \( MT\perp PC \);
\( MT\subset (MPC) \) si \( AT\subset(ABC) \Rightarrow \angle MTA=30 \) ; \( MT=2a \Rightarrow AT=a\sqrt{3} \)
In \( \Delta ACT \) ; \( sin \angle ACT =\frac{sqrt{3}}{2} \Rightarrow \angle ACT=60 \Rightarrow \) In \( \Delta APC \) ; \( tg60=\frac{AP}{2a} \Rightarrow AP=2a\sqrt{3} \Rightarrow PB=BA \Rightarrow BN=\frac{a}{2} \)
\( A_{\Delta MNB}=\frac{a^2\sqrt{3}}{4} \)
\( CA\perp AM \) si \( CA\perp AB \Rightarrow CA \perp (MNB) \)
\( V_{CMNB}=\frac{a^3\sqrt{3}}{6}=\frac{1}{3}\cdot A_{\Delta MNC} \cdot d \Rightarrow d=\frac{a\sqrt{3}}{4} \) .
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