Answer
The magnitude of $B$ is 28.0 m.
Work Step by Step
We can find the angle $\phi$ between the two vectors.
$\phi = (90.0^{\circ}-39.0^{\circ}) + 28.0^{\circ} = 79.0^{\circ}$
We know that the scalar product $\vec{A} \cdot \vec{B} = AB~cos(\phi)$, where $\phi$ is the angle between the two vectors.
$\vec{A} \cdot \vec{B} = AB~cos(\phi) = 48.0~m^2$
$B = \frac{48.0~m^2}{A~cos(\phi)} = \frac{48.0~m^2}{(9.00~m)~cos(79.0^{\circ})}$
$B = 28.0~m$
The magnitude of $B$ is 28.0 m.
