Respuesta :
1)
First we go found the value of [tex]\vec{A}[/tex]:
a² = (-3)² + 7²
a² = 9 + 49
a² = 58
a = √59
a = 7,68 u
Now we can found the angle with the +x-axis
ax = a · cos θ
3 = 7,68 · cos θ
3 / 7,68 = cos θ
0,39 = cos θ
0,39 · [tex]\mathsf{cos^{-1}}[/tex] = θ
θ = 67º <----- This is the answer
2)
First we go found the value of [tex]\vec{B}[/tex]:
b² = 7² + 2²
b² = 49 + 4
b² = 53
b = √53
b = 7,28 u
Now we can found the angle with the +x-axis
bx = b · cos θ
7 = 7,28 · cos θ
7 / 7,28 = cos θ
0,96 · [tex]\mathsf{cos^{-1}}[/tex] = θ
θ = 16º
For counterclockwise:
180 - 16 = 164º <----- This is the answer
3)
[tex]\vec{C}=\vec{A}+\vec{B}[/tex]
[tex]\vec{C}=(-3\vec{i}+7\vec{j})+(7\vec{i}+2\vec{j})[/tex]
[tex]\vec{C}=(-3\vec{i}+7\vec{i})+(7\vec{j}+2\vec{j})[/tex]
[tex] \vec{C}= 4\vec{i}+9\vec{j}[/tex]
Now we going to found the module of C vector:
c² = 9² + 4²
c² = 81 + 16
c² = 97
c = √97
c = 9,84 u
The angle is:
Cx = C · cos θ
4 = 9,84 · cos θ
4 / 9,84 = cos θ
0,40 · [tex]\mathsf{cos^{-1}}[/tex] = θ
θ = 66º
For counterclockwise:
180 - 66 = 114º <----- This is the answer
I hope you enjoy!

The Pythagorean theorem can help to determine the vector's magnitude. The magnitude and angle of the given vector are 7.16 m and 67º respectively.
The magnitude of the vector:
The magnitude of the vector can be calculated by the Pythagorean theorem:
[tex]a =\sqrt { (-3)^2 + 7^2}\\\\ a = \sqrt {58}\\\\ a = 7.16 [/tex]
The angle of a resulting vector can be determined by the formula:
[tex]cos \Theta = \dfrac a{a_x} [/tex]
Here,
[tex]\Theta [/tex] - the angle of resulting vector,
So,
[tex]\rm cos \Theta = \dfrac 3{7.16}\\\\ \Theta = 67^o [/tex]
Therefore, the magnitude and angle of the given vector are 7.16 m and 67º respectively.
Learn more about vector calculation:
https://brainly.com/question/10151539