Question

Jeremy is walking through the woods, marking diseased trees for removal. He walks for 2.44 km, at a heading of 13.9° west of north. Based on this, how far north and how far west has he walked from his starting position?

Answer 1

We conclude that Jeremy walked 2.37km North and 0.59 km West.

How to get the North and West components?

We can assume that the distance that he walked is the hypotenuse of a right triangle, like the one you can see in the image below.

There, you can see that the adjacent cathetus would be the displacement in the North direction, while the opposite cathetus would be the displacement in the West direction.

Then we can use the relations.

• Sin(θ) = (opposite cathetus)/(hypotenuse)
• Cos(θ)  = (adacent cathetus)/(hypotenuse).

With θ = 13.9° and hypotenuse = 2.44km

Solving these two we will get:

sin(13.9°) = Y/2.44km

sin(13.9°)*2.44km = Y = 0.59 km

cos(13.9°) = X/2.44km

cos(13.9°)*2.44km = X = 2.37km

So we can conclude that he walked 2.37km North and 0.59 km West.

If you want to learn more about right triangles, you can read:

brainly.com/question/2217700