Question

1. christopher ran 16 miles at 50 degrees south of west. how far south did christopher run?

-12.3 miles south
16 miles south
12.3 miles south
10.3 miles south

Explanation:

Step1: Analyze the vector components

We can consider Christopher's run as a vector. The distance he ran is the magnitude of the vector ($r = 16$ miles), and the angle south of west is $\theta=50^{\circ}$. To find the south - ward component ($y$ - component), we use the formula $y = r\sin\theta$.

Step2: Calculate the south - ward distance

Substitute $r = 16$ and $\theta = 50^{\circ}$ into the formula $y=r\sin\theta$. We know that $\sin(50^{\circ})\approx0.766$. Then $y = 16\times\sin(50^{\circ})=16\times0.766 = 12.256\approx12.3$ miles. The negative sign is not needed here as we are just asked for the distance (a non - negative quantity representing how far south).

Answer:

12.3 miles south