Chapter 2.4, Problem 240E

$\left[\text{T}\right]$ The fence vector $\text{F}$ acting on a proton with an electric charge of $\text{1}{\text{.6×10}}^{\text{-19}}C$ moving in a magnetic ?eld $B$ where the velocity vector $v$ is given by $F=1.6\times {10}^{-19}\left(v\times B\right)$ (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 $5.9\times {10}^{-17}N$ and the proton is moving at the speed of $300\text{m/sec}$ in magnetic ?eld $\text{B}$ of magnitude $2.4\text{T}$ , find the angle between velocity vector $\text{v}$ of the proton and magnetic ?eld $\text{B}$ . Express the answer in degrees rounded to the nearest integer.