Answer
Angle between force and displacement $=62.3^{\circ}$
Work Step by Step
The magnitude of displacement is:
$=\sqrt {2^{2}+4^{2}+3^{2}}=\sqrt 29 m$
We know, work done is equal to change in Kinetic Energy, so: $Work$ $Done$ $=30$ $J$
We also know that:
$Work$ $Done = F dcos\theta$
Putting the values and solving:
$30= 12\times \sqrt 29 \times cos\theta$
$cos \theta= \frac{30}{64.62} = 0.46$
$\theta =cos^{-1}(0.46) $
$\theta =62.3^{\circ} $
