The figure below represents the observer, train and crossing.

The speed of the observer,

$u=120 \cos \theta$

$u=120\times \frac{288}{\sqrt{{384}^2+{288}^2}}$

$u=120\times \frac{3}{5} \mathrm{km} {\mathrm{h}}^{-1}$

$u=120\times \frac{3}{5}\times \frac{5}{18} \mathrm{m} {\mathrm{s}}^{-1}$

$u=20 \mathrm{m} {\mathrm{s}}^{-1}$

The formula for the frequency heard by observer, when a
sound source is moving towards stationary observer,

$f^{\text{'}}=\left(\frac{v}{v-u}\right) f$

$f^{\text{'}}=\left(\frac{340}{340-20}\right) 576$

$f^{\text{'}}=612 \mathrm{Hz}$