Answer:
The directional bearing of the boat is N 30º E
Step-by-step explanation:
Let [tex]\vec v = (25, 25\sqrt{3})[/tex], where [tex]\vec v[/tex] is the vector velocity. Given that such vector is represented in rectangular, a positive value in the first component is the value of the vector in the east direction, whereas a positive value in the second component is in the north direction. The directional bearing of the boat ([tex]\theta[/tex]), measured in sexagesimal degrees, is determined by trigonometrical means:
[tex]\theta = \tan^{-1}\frac{v_{y}}{v_{x}}[/tex] (1)
If we know that [tex]v_{x} = 25[/tex] and [tex]v_{y} = 25\sqrt{3}[/tex], then the directional bearing of the boat is:
[tex]\theta = \tan^{-1} \sqrt{3}[/tex]
[tex]\theta = 60^{\circ}[/tex]
In consequence, we conclude that the direction bearing of the boat is 30 degrees to the East from the North (N 30º E).
