## Solution:

### Part A

Consider the following figure:

The east distance is the component in the horizontal direction.

$D_E=7.50\:km\:\cdot sin\:\left(15^{\circ} \right)$

$D_E=1.94\:km$

The north distance is the vertical component

$D_E=7.50\:km\cdot cos\left(15^{\circ} \right)$

$D_E=7.24\:km$

### Part B

Based from the figure, we can easily see that the order is reversible in the addition of vectors. We say that $D_E+D_N=D_N+D_E$

## Solution:

Consider the right triangle formed by the legs A, B, and R. We know that A is 18 m, B is 25 m, and we are solving for the magnitude of R. We can do this by using the Pythagorean Theorem. That is

$R=\sqrt{A^2+B^2}$

$R=\sqrt{\left(18\:m\right)^2+\left(25\:m\right)^2}$

$R=\sqrt{324+625}$

$R=\sqrt{949}$

$R=30.8\:m$

So, the distance is about 30.8 meters from the starting point. To solve for the value of the unknown angle, φ, we can use the tangent function. That is

$tan\:\phi =\frac{A}{B}$

$tan\:\phi =\frac{18\:m}{25\:m}$

$\phi =tan^{-1}\left(\frac{18}{25}\right)$

$\phi =35.75^{\circ}$

Therefore, the compass angle is $35.75^{\circ} \:West\:of\:North$