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.

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The second equation should be $\sin\beta=-\dfrac{\sin\alpha}{1.341}$. –  Daryl Dec 3 '12 at 23:17
Something is not correct because $0 = \sin\alpha + 1.341\sin\beta$ is equivalent to $\sin\beta = -\frac{\sin\alpha}{1.341}$. –  Américo Tavares Dec 3 '12 at 23:17
Sorry, that was a typo. I will correct it. –  Dylan Dec 3 '12 at 23:18
The minus sign is still missing. –  Américo Tavares Dec 3 '12 at 23:52
Sorry, the sign in the original equation $0 = \sin\alpha + 1.341\sin\beta$ should have read $0 = \sin\alpha - 1.341\sin\beta$ –  Dylan Dec 4 '12 at 6:16

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 $\beta$.

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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 $\displaystyle \alpha =-\arccos\frac{k^2-5}4$.

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