Answer
See explanation.
Work Step by Step
a. The magnitude is $\sqrt{A_x^2+ A_y^2}= 10.0$ cm.
The angle is $tan^{-1}(\frac{A_y}{A_x})=149^{\circ}$. The vector is in the second quadrant.
b. The magnitude is $\sqrt{A_x^2+ A_y^2}= 10.0$ m.
The angle is $tan^{-1}(\frac{A_y}{A_x})=194^{\circ}$. The vector is in the third quadrant.
c. The magnitude is $\sqrt{A_x^2+ A_y^2}= 8.21$ km.
The angle is $tan^{-1}(\frac{A_y}{A_x})=-19.2^{\circ}$. The vector is in the fourth quadrant.
