The problem is as follows:
The figure from below shows a hydraulic cylinder held to the ground which is connected to two arms $AB$ and $BC$. In the instant that $\alpha=\beta=60^{\circ}$ the cylinder transmits to $A$ a speed of $v_{A}=1.2\sqrt{3}\,\frac{m}{s}$ (to the left). Find the angular speed in $\frac{rad}{s}$ of the bars $BC$ and $AB$.
The alternatives in my book are as follows:
$\begin{array}{ll} 1.&4.8\frac{rad}{s}\,\textrm{and}\,2.4\frac{rad}{s}\\ 2.&2.4\frac{rad}{s}\,\textrm{and}\,4.8\frac{rad}{s}\\ 3.&2.4\frac{rad}{s}\,\textrm{and}\,1.2\frac{rad}{s}\\ 4.&3.6\frac{rad}{s}\,\textrm{and}\,1.8\frac{rad}{s}\\ 5.&4.8\frac{rad}{s}\,\textrm{and}\,3.6\frac{rad}{s}\\ \end{array}$
I'm totally lost in this question. Can somebody instruct me exactly how am I supposed to find the angular speed?.
The only thing which I can barely remember is that the torque is related with the angular acceleration but I'm not sure if this will be applied here. Can an answer be detailed the most as possible so I can get the idea of what's going on here?.
Denote by $(i,j)$ a fixed orthonormal frame where $i$ is horizontal and $j$ vertical. You have following equations for the speeds of points $A, B, C$ using relative speed formula:
$$\begin{cases} v_A &= -(1.2 \sqrt{3})i\\ v_B &= v_A + \dot{\alpha}AB(- (\sin \alpha) i + (\cos \alpha) j)\\ v_B &= (\dot{\alpha}+\dot{\beta})BC(- \sin (\alpha+\beta) i + \cos (\alpha+\beta) j) \end{cases}$$ as $v_C =0$.
Therefore $$-(1.2 \sqrt{3})i+\dot{\alpha}AB(- (\sin \alpha) i + (\cos \alpha) j)-(\dot{\alpha}+\dot{\beta})BC(- \sin (\alpha+\beta) i + \cos (\alpha+\beta)=0$$
Based on that you get the two real equations (one for each coordinate):
$$\begin{cases} 1.2 \sqrt{3} + \dot{\alpha}AB\sin \alpha- (\dot{\alpha}+\dot{\beta})BC\sin (\alpha+\beta) &=0\\ \dot{\alpha}AB\cos \alpha - (\dot{\alpha}+\dot{\beta})BC\cos (\alpha+\beta) &=0 \end{cases}$$
From which you can get $\dot{\alpha}, \dot{\beta}$.