$\vec{u} = (-9,3)$ $\newline$ Find the direction angle of $\vec{u}$ . $\newline$ Enter your answer as an angle in degrees between $0^{\circ}$ and $360^{\circ}$ rounded to the nearest hundredth. $\newline$ $\theta=$ $\square^{\circ}$

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\vec{u} = (-9,3)

\newline

Find the direction angle of

\vec{u}

\newline

Enter your answer as an angle in degrees between

0^{\circ}

and

360^{\circ}

rounded to the nearest hundredth.

\newline

\theta=

\square^{\circ}

Calculate tangent: To find the direction angle of $\vec{u}$ , we need to use the arctangent function, which gives us the angle whose tangent is the quotient of the y-coordinate and the x-coordinate of the vector.
Find arctangent: First, let's calculate the tangent of the direction angle, which is the y-coordinate divided by the x-coordinate of $\vec{u}$ . So, $\tan(\theta) = \frac{3}{-9}$ .
Calculate angle: Now, we calculate $\tan(\theta) = \frac{3}{(-9)} = -\frac{1}{3}$ .
Adjust for quadrant: Next, we use the arctangent function to find the angle. So, $\theta = \text{arctan}(-\frac{1}{3})$ . We'll use a calculator for this.
Adjust for quadrant: Next, we use the arctangent function to find the angle. So, $\theta = \text{arctan}(-\frac{1}{3})$ . We'll use a calculator for this.After using the calculator, we find that $\theta \approx \text{arctan}(-\frac{1}{3}) \approx -18.43^\circ$ .
Adjust for quadrant: Next, we use the arctangent function to find the angle. So, $\theta = \text{arctan}(-\frac{1}{3})$ . We'll use a calculator for this.After using the calculator, we find that $\theta \approx \text{arctan}(-\frac{1}{3}) \approx -18.43^\circ$ .However, we want the direction angle to be between $0^\circ$ and $360^\circ$ . Since our vector is in the second quadrant (negative $x$ and positive $y$ ), we add $180^\circ$ to the angle we found.
Adjust for quadrant: Next, we use the arctangent function to find the angle. So, $\theta = \text{arctan}(-\frac{1}{3})$ . We'll use a calculator for this.After using the calculator, we find that $\theta \approx \text{arctan}(-\frac{1}{3}) \approx -18.43^\circ$ .However, we want the direction angle to be between $0^\circ$ and $360^\circ$ . Since our vector is in the second quadrant (negative x and positive y), we add $180^\circ$ to the angle we found.So, $\theta = -18.43^\circ + 180^\circ = 161.57^\circ$ .