Question

a vector in standard position has its terminal point at (-10, -4). what is the approximate direction angle of the vector? 22° 58° 158° 202°

Explanation:

Step1: Recall the formula for direction angle

If a vector has terminal - point $(x,y)$ in standard position, the formula for the direction angle $\theta$ is $\tan\theta=\frac{y}{x}$, and we need to adjust the angle based on the quadrant. Here $x = - 10$ and $y=-4$.

Step2: Calculate the reference angle

First, find the magnitude of $\tan\theta$. $\tan\theta=\frac{y}{x}=\frac{-4}{-10}=\frac{2}{5}=0.4$. Then, the reference angle $\theta_{r}=\arctan(0.4)\approx22^{\circ}$.

Step3: Determine the quadrant and the direction angle

Since $x=-10\lt0$ and $y = - 4\lt0$, the vector lies in the third - quadrant. The direction angle $\theta$ in the third - quadrant is given by $\theta = 180^{\circ}+\theta_{r}$. So $\theta=180^{\circ}+22^{\circ}=202^{\circ}$.

Answer:

D. $202^{\circ}$