Răspuns:
B-C-A-D
Explicație pas cu pas:
[tex]AB = AC \cdot \sec{30^\circ} = AC\cdot \frac{2}{\sqrt{3}} = AC \cdot \frac{2\sqrt{3}}{3} = 12\cdot \frac{2\sqrt{3}}{3} = 8\sqrt{3}\\\\ BC = \sin{30^\circ}\cdot 8\sqrt{3} = \frac{1}{2}\cdot 8\sqrt{3} = 4\sqrt{3}\\\\CD = AC\cdot \sec{30^\circ} = 8\sqrt{3}\\\\ AD = 8\sqrt{3}\cdot \sin{30^\circ} = 4\sqrt{3}\\\\B\to C\to A \to D = BC + CA + AD = 4\sqrt{3} + 12 + 4\sqrt{3}= 12 + 8\sqrt{3}\\\\ A\to B\to C\to D = AB + BC + CD = 8\sqrt{3} + 4\sqrt{3} + 8\sqrt{3} = 20\sqrt{3}\\\\12+8\sqrt{3} < 20\sqrt{3}\\\\12 < 12\sqrt{3}\\\\ 1 < \sqrt{3}\\\\ 1 < 3 \\\\\implies \textrm{Vaporul care parcurge traseul } B\to C\to A\to D \textrm{ ajunge mai repede la destinatie}\\\\\textrm{daca ambele vapoare merg cu viteza egala}[/tex]
