In a physics problem, I am asked to find the resulting angle of two velocity vectors using each velocity vector's components.
For the x-component, I have $m_av_{1ax} = m_av_{2ax} + m_bv_{2bx}$. Plugging in the given values for this component gives $2 = \cos\alpha + 1.341\cos\beta$. (note that I left out units for mass and velocity vectors, but those should cancel anyway since it is a ratio of mass times velocity)
For the y-component, I have $m_bv_{1by} = m_av_{2ay} + m_bv_{2by}$. Plugging in the given values for this component gives $0 = \sin\alpha - 1.341\sin\beta$
The book recommends solving this with simultaneous equations, but I am not sure how to do this when working with angles. Their hint suggests solving for beta first: $ \cos\beta = \frac{2-\cos\alpha}{1.341}$ $ \sin\beta = \frac{\sin\alpha}{1.341}$ And then using Pythagorean theorem by squaring both $\cos$ and $\sin$: $\cos^2\beta + \sin^2\beta = 1$ However, I am still not clear on how to isolate the angle for $\alpha$ when substituting the above definitions for $\cos\beta$ and $\sin\beta$. Furthermore, plugging the values for both these functions into a matrix and solving them simultaneously doesn't seem to be very useful for finding the angle.