Answer
$A_x = -8.12 ~m$
$A = 15.3~m$
Work Step by Step
We can find the magnitude of $A_x$.
$\frac{A_x}{A_y} = tan(\theta)$
$A_x = (13.0 ~m)~tan(32.0^{\circ}) = 8.12~m$
Since the angle is west of the positive y-axis, then $A_x = -8.12 ~m$.
We can use $A_x$ and $A_y$ to find the magnitude of A.
$A = \sqrt{(A_x)^2 + (A_y)^2}$
$A = \sqrt{(-8.12 ~m)^2 + (13.0 ~m)^2}$
$A = 15.3~m$
