Calculate angle between two lines We have four points: a, b, c and d. We only know length of line cd and line ab. We also know that points c and d have same x coordinate, also points a and b have same x coordinate. Lines cd and ab are parallel. How can I find angle (marked as green) between lines ca and line db? Please provide example of calculations.

 A: Given the points with their coordinates $a_x, a_y ... d_x, d_y$, you can compute the angle of each line to the x-axis (using the arcustangens function), and then just compute the difference between these angles. 
So you can compute the angles 
$\alpha_0 = atan2(c_y-a_y, c_x-a_x)$
and
$\alpha_1 = atan2(d_y-b_y, d_x-b_x)$
(see the wikipedia link for the definition of the $atan2$ function). 
The angle between the lines is then simply $\alpha_1-\alpha_0$

EDIT: The example:


*

*$a_x = 100$, $a_y = 5$

*$b_x = 100$, $b_y = 0$

*$c_x = 0$, $c_y = 8$

*$d_x = 0$, $d_y = 0$


Now compute
$\alpha_0 = atan2(c_y-a_y, c_x-a_x) = atan2(3, -100)$
$\alpha_1 = atan2(d_y-b_y, d_x-b_x) = atan2(0, -100)$
According to the cases described in this image from wikipedia:

We have 
$\alpha_0 = atan(3/-100) + \pi \approx 3.1116016487329152$
$\alpha_1 = atan(0/-100) + \pi \approx 3.141592653589793$
And the angle is then the difference
$\alpha = \alpha_1 - \alpha_0 \approx 0.02999100485687789 \approx 1.718°$
A: If you have two lines given by vector equations, say $\bf{p}+\lambda{\bf u}$ and ${\bf q}+\mu{\bf v}$, and the accute angle between the lines is $\theta$ then there is a well-known relationship:
$$\cos\theta = \left| \frac{{\bf u} \cdot {\bf v}}{\|{\bf u}\|\,\|{\bf v}\|} \right|$$
Notice that only the directions ${\bf u}$ and ${\bf v}$ of the two lines are needed. 
If a line passes through points ${\bf a}$ and ${\bf c}$ then the line has direction ${\bf c}-{\bf a}$ because ${\bf c}-{\bf a}$ is the vector from  ${\bf a}$ to ${\bf c}$. If you know all of the coordinates of the points then you can work out the direction vectors of the two lines and then apply the formula to get $\cos\theta$. Then use $\cos^{-1}$ to get $\theta$.
A: we make a line parellel to the line $bd$, showed in this picture:
then the angle you want is equal to the angle $\angle cad'$, and $dd'=ab$, because $ab\parallel cd$.
so $cd'=cd-ab$.
now we can calculate the angle $\angle cad'$ in the triangle $\bigtriangleup cad'$.
$$\cos(\angle cad')=\frac{ac^2+bd^2-d'c^2}{2\cdot ac\cdot bd}$$
then you have the angle by the function $\arccos()$.
In fact your inital conditions are not enough. from the two following pictures, we know that:
1) when we only know the length of $ab$ and $cd$, the angle between $ac$ and $bd$ depends the distance between $ab$ and $cd$;
2) though $ab$ and $cd$ has a certain distance $\Delta x$, the angle between $ac$ and $bd$ depends also the hight of $ab$.
A: If I understand you question correctly: cd and ab are parallel. So you can shrink your problem to a triangle. Where your requested angle (let's call him $\alpha$) depends on the angles of $cdb$ (here $\beta$) and $dca$ (here $\gamma$) or any other two angles in your figure. 
Thus all these three angles must be $180° = \alpha + \beta + \gamma$ in the sum.
So I think your problem is under determined.
