Explicație:
a)
[tex]AM = \frac{AB \sqrt{3} }{2} =\frac{6\sqrt{3} }{2} = 3 \sqrt{3} [/tex]
[tex]OM = \frac{1}{3} \times AM = \sqrt{3} [/tex]
[tex]VO = \sqrt{VM^{2} - OM^{2} } = \sqrt{ {2}^{2} - 3} = 1[/tex]
[tex]V= \frac{1}{3} \times h \times aria(ABC) = \frac{VO \times AM \times BC}{6} = \frac{1 \times3 \sqrt{3} \times 6}{6} = 3 \sqrt{3} \: {cm}^{3} = \sqrt{27} > \sqrt{25} = 5 \: {cm}^{3}[/tex]
[tex]= > V > 5 \: {cm}^{3}[/tex]
b) VM ⊥ BC și AM ⊥ BC
⇒∡((VBC),(ABC)) = ∡VMO
[tex]\sin(VMO) = \frac{VO}{VM} = \frac{1}{2}\\[/tex]
⇒∡VMO = 30°
