Asailor in a small sailboat encounters shifting winds. she sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. her final position is 5.80 km directly east of the starting point

Explanation:

First, the problem is incomplete, but, in a different site, I found this question. The remaining part is to find the magnitude of the third leg of this journey, which is the unknown here.

In order to do this, we need to know how is the trip. The attached picture shows the road that follows this trip.

You should remember that if you a x-y coordinate system, and states that North, South, East and West are on the axis, the south east direction is half between South and East. And that form an angle of 45° for the further calculations

Now that we have cleared this, let's begin with the exercise.

According to the picture, the first leg is on the x axis heading east, so it does not have y components. So, the distance traveled is 2 km in x, and 0 in y.

For the second leg, let's calculate the component in x and y for this, using the 45°

d2x = 3.5cos45 = 2.47 km

d2y = 3.5sin45 = 2.47 km ---> in the y axis is heading south, that's why this result would be negative so: -2.47 km

For the third leg, we only know that the final position is 5.80 km from the starting point, so the x component of all distance should be summed and the result has to be 5.80 therefore:

5.80 = d1x + d2x + d3x

From here, solve for d3x:

5.80 - 2 - 2.47 = d3x

d3x = 1.33 km

We do the same for the y components. In this case, the sum of the y components would be 0, so:

0 = d1y + d2y + d3y

0 = 0 - 2.47 + d3y

d3y = 2.47 km

To get the magnitude of this leg, we apply the next expression:

d = √(dx² + dy²)

Solving for d3:

d3 = √(1.33)² + (2.47)²

d3 = 2.81 km

This would be the magnitude of the third leg.

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