Answer
(a) $A_y =−23.7~ m$
(b) $A=28.6 ~m$
Work Step by Step
(a) We can find the magnitude of $A_y$.
$\frac{A_x}{A_y}= tan(\theta)$
$A_y= \frac{A_x}{tan(\theta)} = \frac{16.0~m}{tan(34^{\circ})} = 23.7 ~m$
Since the angle is clockwise from the negative y-axis, then
$A_y =−23.7~ m$
(b) We can use $A_x$ and $A_y$ to find the magnitude of $A$.
$A^2=(A_x)^2+(A_y)^2$
$A=\sqrt{(−16.0 ~m)^2+(-23.7 ~m)^2}$
$A=28.6 ~m$
