A person walks 15.0° north of east for 3.70 km. How far due north and how far due east would she have to walk to arrive at the same location?

It's easiest to just draw a diagram.

Start out at (0,0) and draw a line 3.7 long, at an angle of 15deg up from horizontal.

Now you have a right triangle, with hypotenuse 3.7, and
height = 3.7 sin 15 = 0.96 (north)
base = 3.7 cos 15 = 3.57 (east)

To find the distance walked due north and due east, we can use trigonometry.

Let's start by drawing a diagram to visualize the problem:

```
|
E | X
|
-----+-----
|
|
|
|
N |
```

Here, the horizontal axis represents east, and the vertical axis represents north. The person starts at the origin and walks 15.0° north of east for 3.70 km, ending up at point X.

Now, let's break down the motion into its north and east components. We can do this by using trigonometry and the given angle:

The north component is the vertical segment from X to the north axis. Let's call its length "dN".

The east component is the horizontal segment from X to the east axis. Let's call its length "dE".

To find the values of dN and dE, we can use trigonometric functions. Since the person walks 15.0° north of east, it forms a right-angled triangle with the north and east axes.

The opposite side of the triangle is dN, the adjacent side is dE, and the hypotenuse is the distance walked (3.70 km).

Using trigonometry, we can write:

sin(15.0°) = dN / 3.70 km

cos(15.0°) = dE / 3.70 km

To solve for dN and dE, we can rearrange the equations:

dN = sin(15.0°) * 3.70 km

dE = cos(15.0°) * 3.70 km

Now, let's calculate the values:

dN = sin(15.0°) * 3.70 km
dN ≈ 0.254 * 3.70 km
dN ≈ 0.940 km

dE = cos(15.0°) * 3.70 km
dE ≈ 0.966 * 3.70 km
dE ≈ 3.578 km

So, to arrive at the same location, the person would need to walk approximately 0.940 km due north and 3.578 km due east.