A proton travels through a region of uniform magnetic field at an angle θ relative to the magnetic field. The magnitude of the magnetic field in this region is 1.15 T, and the proton's velocity is 500,000 m/s when it experiences an acceleration whose magnitude is 1.57 x10^{13} m/s^2. Calculate the angle θ. Give your answer in degrees.
$Massofproton\left(m\right)=1.67\times {10}^{-27}kg\phantom{\rule{0ex}{0ex}}Chargeofproton\left(q\right)=1.6\times {10}^{-19}C\hspace{0.17em}\phantom{\rule{0ex}{0ex}}Magneticfieldstrength\left(B\right)=1.15T\hspace{0.17em}\phantom{\rule{0ex}{0ex}}Velocityofproton\left(v\right)\hspace{0.17em}=500000\frac{m}{s}\phantom{\rule{0ex}{0ex}}Accelerationofproton\left(a\right)\hspace{0.17em}=1.57\times {10}^{13}\frac{m}{{s}^{2}}\phantom{\rule{0ex}{0ex}}$
$U\mathrm{sin}gtheformulaformagneticforce:\phantom{\rule{0ex}{0ex}}F\hspace{0.17em}=q\times v\times B\times Sin\theta \phantom{\rule{0ex}{0ex}}U\mathrm{sin}gnewton\text{'}ssecondlawofmotion:\phantom{\rule{0ex}{0ex}}netforce=mass\times acceleration\phantom{\rule{0ex}{0ex}}mass\times acceleration=q\times v\times B\times Sin\theta \phantom{\rule{0ex}{0ex}}a=\frac{q\times v\times B\times Sin\theta}{m}\phantom{\rule{0ex}{0ex}}1.57\times {10}^{13}=\frac{1.6\times {10}^{-19}\times 500000\times 1.15\times Sin\theta}{1.67\times {10}^{-27}}\phantom{\rule{0ex}{0ex}}1.57\times {10}^{13}=5.508\times {10}^{13}\times Sin\theta \phantom{\rule{0ex}{0ex}}0.285=Sin\theta \phantom{\rule{0ex}{0ex}}16.{55}^{o}=\theta \phantom{\rule{0ex}{0ex}}(Anglebetweenvelocityandmagneticfield)$