Question

A boat travels with velocity vector (25, 25 StartRoot 3 EndRoot). What is the directional bearing of the boat? N 30° E E 30° S E 30° N N 30° W

Answer 1

Answer:

a

Step-by-step explanation:

Answer 2

Answer:

The directional bearing of the boat is N 30º E

Step-by-step explanation:

Let $\vec v = (25, 25\sqrt{3})$, where $\vec v$ 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 ($\theta$), measured in sexagesimal degrees, is determined by trigonometrical means:

$\theta = \tan^{-1}\frac{v_{y}}{v_{x}}$ (1)

If we know that $v_{x} = 25$ and $v_{y} = 25\sqrt{3}$, then the directional bearing of the boat is:

$\theta = \tan^{-1} \sqrt{3}$

$\theta = 60^{\circ}$

In consequence, we conclude that the direction bearing of the boat is 30 degrees to the East from the North (N 30º E).