Explanation
a vectors makes a rigth angle to the x-positve axis , so
so, the x coordinate is adjacent side and the y-coordinate becomes into the opposite side, then we can use a trigonometric function that relates those values,it is
[tex]\begin{gathered} tan\theta=\frac{opposi\text{te side}}{adjacent\text{ side}} \\ tan\theta=\frac{y\text{ coordinate}}{x\text{ coordinate}} \end{gathered}[/tex]
hence
Step 1
a)
let
[tex]\begin{gathered} \langle5,3\rangle, \\ x=5 \\ y=3 \end{gathered}[/tex]
replace and solve for the angle
[tex]\begin{gathered} tan\theta=\frac{y\text{coord\imaginaryI nate}}{x\text{coord\imaginaryI nate}} \\ tan\theta=\frac{3}{5} \\ \theta=\tan^{-1}(\frac{3}{5}) \\ \theta=30.964° \end{gathered}[/tex]
so,
a)Blank1: 30.964
Step 2
b)
let
[tex]\begin{gathered} \langle-4,5\rangle, \\ x=-4 \\ y=5 \end{gathered}[/tex]
replace and solve for the angle
[tex]\begin{gathered} tan\theta=\frac{y\text{coord\imaginaryI nate}}{x\text{coord\imaginaryI nate}} \\ tan\theta=\frac{5}{-4} \\ \theta=\tan^{-1}(\frac{5}{-4}) \\ \theta=-51.340\~+180(Iquadrant) \\ \theta=128.660 \\ . \end{gathered}[/tex]
so,
b)Blank2:128.660
Step 3
c)
[tex]\begin{gathered} \langle8,-8\rangle, \\ x=8 \\ y=-8 \end{gathered}[/tex]
replace and solve for the angle
[tex]\begin{gathered} tan\theta=\frac{y\text{coord\imaginaryI nate}}{x\text{coord\imaginaryI nate}} \\ tan\theta=\frac{-8}{8} \\ \theta=\tan^{-1}(-1) \\ \theta=-45 \end{gathered}[/tex]
so,
c)Blank3:-45 °
Step 4
d)
[tex]\begin{gathered} \langle-12,-3\rangle, \\ x=-12 \\ y=-3 \end{gathered}[/tex]
replace and solve for the angle
[tex]\begin{gathered} tan\theta=\frac{y\text{coord\imaginaryI nate}}{x\text{coord\imaginaryI nate}} \\ tan\theta=\frac{-3}{-12} \\ \theta=\tan^{-1}(\frac{1}{4}) \\ \theta=14.036 \end{gathered}[/tex]
so,
direction
[tex]\begin{gathered} direcgtion\text{ =}\theta+180=14.036 \\ angle=194.036 \end{gathered}[/tex]
graph
d)Blank4: 194.036
I hope this helps you
