A person walks 28.2◦ north of east for 3.33 km.

Another person walks due north, then due
east to arrive at the same location.
How far due north would this person walk?
How far would this person walk due east?

how far N? 3.33km*sin28.2

how far E? 3.33km*cos28.2

draw the figure, and verify this.

first understand d question

draw d diagram
and

To solve this problem, we can use trigonometry.

First, let's consider the person who walks 28.2◦ north of east for 3.33 km. This can be represented as a vector with a magnitude of 3.33 km, pointing 28.2◦ north of east.

To find the northward component of this vector, we can use the sine function. The formula is:

Northward component = Magnitude * sin(angle)

Substituting the given values, we have:

Northward component = 3.33 km * sin(28.2◦)

Calculating this, we find that the northward component is approximately 1.531 km.

Next, to find the eastward component of the vector, we can use the cosine function. The formula is:

Eastward component = Magnitude * cos(angle)

Substituting the given values, we have:

Eastward component = 3.33 km * cos(28.2◦)

Calculating this, we find that the eastward component is approximately 2.856 km.

Now, let's consider the person who walks due north, then due east to arrive at the same location. Since both components are perpendicular, we can use the Pythagorean theorem to find the total distance traveled.

The Pythagorean theorem states:

Total distance = √(Northward component^2 + Eastward component^2)

Substituting the values we calculated earlier, we have:

Total distance = √(1.531 km^2 + 2.856 km^2)

Calculating this, we find that the total distance is approximately 3.255 km.

Since the person walks due north and due east, the distance due north and due east will be equal.

Thus, the person would walk approximately 1.628 km due north and approximately 1.628 km due east.