The problem is as follows:
The figure from below shows an equilateral triangle $\triangle=EBC$ and $\angle\,EDC=70^{\circ}$. Using this condition find $x^{\circ}$.
The alternatives given in my book are as follows:
$\begin{array}{ll} 1.&30^{\circ}\\ 2.&15^{\circ}\\ 3.&20^{\circ}\\ 4.&25^{\circ}\\ 5.&35^{\circ}\\ \end{array}$
This problem has left me going in circles. Can someone help me here?.
The only thing which I could spot was $EB=BC=EC$ and $\angle\,EAC=20^{\circ}$
But that's it. I don't know what else can be done to identify or spot the required angle. Does this thing requires some sort of construction or any identity such as congruence or something?. I just can't find any.
Please include a drawing in the answer or sketch to spot where to look for, as I am not that good to identifying the sides and angles from only looking algebraic expressions.
As you requested, the following is your figure with several extra degree values being specified, and congruent side lengths marked with a short line segment roughly perpendicular to the side.
Since $\measuredangle EDC = 70^{\circ}$, then the sum of angles in $\triangle ECD$ being $180^{\circ}$ means $\measuredangle CED = 180^{\circ} - 70^{\circ} - 40^{\circ} = 70^{\circ}$. Thus, $\triangle ECD$ is isosceles (as Brian Tung's question comment indicated), with $\left| EC \right| = \left| DC \right|$.
The stated detail of $\triangle EBC$ being equilateral means $\left| EC \right| = \left| BC \right|$, giving $\left| DC \right| = \left| BC \right|$, so $\triangle BCD$ is also isosceles. With $\measuredangle BCD = 40^{\circ} + 60^{\circ} = 100^{\circ}$, this means $\measuredangle CBD = \measuredangle CDB = \frac{180^{\circ} - 100^{\circ}}{2} = 40^{\circ}$. Finally, $x = \measuredangle EDC - \measuredangle CDB = 70^{\circ} - 40^{\circ} = 30^{\circ}$, i.e., alternative $1$.