$[T]$ The force vector $F$ acting on a proton with anelectric charge of $1.6 \times 10^{-19} \mathrm{C}$ moving in a magnetic
field $B$ where the velocity vector $V$ is given by$\mathbf{F}=1.6 \times 10^{-19}(\mathbf{v} \times \mathbf{B})$ (here, v is expressed in meters
per second, $B$ in $T$, and $F$ in $N$). If the magnitude of force $F$ acting on a proton is$5.9 \times 10^{-17} \mathrm{~N}$ and the
proton is moving at the speed of 300 m/sec in magnetic field $B$ of magnitude $2.4 T$, find the angle between velocity
vector $V$ of the proton and magnetic field $B$. Express the answer in degrees rounded to the nearest integer.