How to find the angle of emission of an X-ray during impact of accelerated cathode rays?
The angle of resulting X-ray radiation, product of the impact, is related to the amount of energy or momentum of the cathode ray. Bremsstrahlung relation will amount how much energy the produced photon has.
$$hf = \frac{hc}{\lambda} = K - K'\tag{1}$$
$$\lambda_\mathrm{min} = \frac{hc}{K} = \frac{hc}{e\,\Delta V}\tag{2}$$
Back scatter the possible photon angles with the Compton effect equation,
$$\lambda' - \lambda = \frac{h}{m_\mathrm{e}c}(1 - \cos\theta)\tag{3}$$
A difference in kinetic energy is as difference in wavelength. Precise results should consider a vacuum and use imaginary numbers; also probability distribution on superposition.
For more details refer to the chapter 3 of Krane's Modern Physics [1].
Reference

Krane, K. S. Modern Physics, 3rd ed.; Wiley: Hoboken, NJ, 2012. ISBN 978-1-118-06114-5.