Angle between two lines explanation I gave two lines $q_1 = 2x - y + 2 = 0$ and $q_2 = x + 2y - 3 = 0$. 
They have vectors $n_1 = (2,1)$ and $n_2 = (1,2)$. 
When I have to find angle between them I must apply that formula: 
$$\cos \theta = \frac{2\cdot 1 + (-1)\cdot2}{\sqrt{4 + 1}\sqrt{1 + 4}} = 0 \Rightarrow \theta =  90^\circ$$
Replacing $n_1$ and $n_2$ is clear to me, but can you please explain how I get $90^\circ$ from solving the formula. Thank you.
Where those $\sqrt{4 + 1}\sqrt{1 + 4}$ comes from?
If $ \theta $ is $90^ \circ$ at $0$, how much degrees it will be if it is $\dfrac{1}{5}$?
 A: Given two vectors in $\mathbb{R}^2$, say $(x_1, y_1)$ and $(x_2, y_2)$, we have the formula 
$$\cos \theta = \frac{(x_1, y_1) \cdot (x_2, y_2)}{\|(x_1, y_1)\| \|(x_2,y_2)\|}.$$
Some clarification about the used notation: $\cdot$ denotes the inproduct of two vectors. It is defined as follows:
$$(x_1,y_1) \cdot (x_2,y_2) = x_1x_2 + y_1y_2.$$
The norm of the vector $(x_1,y_1)$ is defined as $$\|(x_1, y_1\| = \sqrt{(x_1,y_1) \cdot (x_1,y_1)} = \sqrt{x_1^2 + y_1^2}.$$
This is the length of the vector and if you would draw the vector in the plane and draw the projections to the x and y axis, you'll see that this just is Pythagoras' theorem. 
You know that if $\cos \theta = 0$, then we have that $\theta = 90^{\circ}$. 
The square roots come from the formula I gave you (and you probably find in your coursenotes).
If you want to find $\theta$ where $\cos \theta$ = 1/5, you just compute the arccosine of 1/5.
A: Let $\theta$ be an angle.
$\cos \theta = \frac{2.1 + (-1).2}{\sqrt 5. \sqrt 5}$
$\cos \theta = \frac{2 - 2}{\sqrt 5. \sqrt 5}$
$\cos \theta = \frac{0}{5}$
$\cos \theta = 0$
$\cos \theta = \cos 90°$
$\theta = 90°$
