Finding angles of two vectors using simultaneous equations 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.
 A: I assume that you need to solve the following system
$$
\left\{ 
\begin{array}{c}
2=\cos \alpha +1.341\cos \beta  \\ 
0=\sin \alpha +1.341\sin \beta 
\end{array}
\right.$$ 
Hint. Correct your 2nd. equation
$$ \sin\beta = -\frac{\sin\alpha}{1.341}$$
and notice that
$$
\left( \frac{2-\cos \alpha }{1.341}\right) ^{2}+\left( -\frac{\sin \alpha }{1.341}\right) ^{2}=1
$$
is equivalent to
$$
4-4\cos \alpha +\cos ^{2}\alpha +\sin ^{2}\alpha =1.341^{2}.
$$
Then solve for $\alpha$ and use the result to evaluate $\beta$.
A: Let $k=1.341$ and follow the hint:
 Solve for beta and insert
$\beta=\arcsin\left(\frac{\sin \alpha}{k}\right) $ into the first equation to get:
$$
\cos\arcsin\left(\frac{\sin \alpha}{k}\right)=\sqrt{1-\frac{\sin^2\alpha}{k^2}} = \frac{2-\cos\alpha}{k}
$$
Square it to get 
$$
\begin{eqnarray}
1&-&\frac{\sin^2\alpha}{k^2}=\left(\frac{2-\cos\alpha}{k}\right)^2=\frac4{k^2} -\frac{4\cos\alpha}{k^2} + \frac{\cos^2\alpha}{k^2}\\
1&=&\frac4{k^2} -\frac{4\cos\alpha}{k^2} + 
\frac{\cos^2\alpha}{k^2}+\frac{\sin^2\alpha}{k^2}\\
&&\phantom{k^2=3-4\cos\alpha}\\
&&\phantom{k^2=4-4\cos\alpha}\\
k^2&=&5-4\cos\alpha.
\end{eqnarray}
$$
You'll get $\displaystyle \alpha =-\arccos\frac{k^2-5}4$.
