In the fig., the red line has a slope of $\frac{1}{2}$ and the green line has a slope of 3.
We can find the angle $\theta$ between the lines as follows:
$\begin{array}{ll} {}&{\tan \theta} &{}={}& {\frac{m_2~-~m_1}{1~+~m_1 m_2}} &{} \\ {}&{} &{}={}& {\frac{3~-~\frac{1}{2}}{1~+~\frac{1}{2} × 3}} &{} \\ {}&{} &{}={}& {\frac{6 ~-~1}{2~+~3}} &{} \\ {}&{} &{}={}& {\frac{5}{5}} &{} \\ {}&{} &{}={}& {1} &{} \\ \end{array}$
So we have to solve the equation $\tan \theta = 1$.
But there are two values that will satisfy the equation. They are: $45^o~\text{and}~225^o$.
Could you please explain how $225^o$ is related to the "angle between the two lines". Thanks.
Let $A,B,C,D$ denote the points $(2,1.5), (4,2.5), (2.5,3),$ and $(1.5,0).$
Then, the red and green lines intersect at point $A$, point $B$ is on the red line and both of points $C$ and $D$ are on the green line.
Then, $\angle BAC = 45^\circ$ and $\angle BAD = 225^\circ,$ when traversed in a counter-clockwise direction.