Answer
$28m, 19m$
Work Step by Step
We can find the required magnitudes of $\vec A$ and $\vec B$ as follows:
We know that
$A_y+B_y+C_y=0$
$\implies B_y=-C_y-A_y=-(C_ysin40.0^{\circ})-0$
$\implies B_y=(15m)sin40.0^{\circ}=9.64m$
Now $B_x$ can be determined as:
$tan 30.0^{\circ}=\frac{B_y}{B_x}$
$\implies B_x=\frac{B_y}{tan 30.0^{\circ}}=\frac{9.64}{tan 30.0^{\circ}}=16.7m$
Now we find $A_x$
$A_x+B_x+C_x=0$
$\implies A_x=-B_x-C_x=-(16.7)-(15)cos40.0^{\circ}=-28m$
The magnitude of $\vec A$ is given as
$|\vec A|=\sqrt{A_x^2+A_y^2}=\sqrt{(-28)^2+(0)^2}=28m$
Now the magnitude of $\vec B$ is given as
$|\vec B|=\sqrt{(B_x)^2+(B_y)^2}=\sqrt{(16.7)^2+(9.64)^2}=19m$
