find the acute angle between the lines:
x - 2y + 1 = 0
and
y = 5x - 4
1. $\displaystyle x - 2y + 1 = 0 ~\implies~y=\frac12 x +\frac12$
$\displaystyle y = 5x - 4$
2. The angle between the first line and the horizontal is $\displaystyle \alpha$. Then you know that $\displaystyle \tan(\alpha) = \frac12$ .
The angle between the second line and the horizontal is $\displaystyle \beta$. Then you know that $\displaystyle \tan(\beta) = 5$ .
3. The angle which is included by the 2 lines is $\displaystyle (\beta - \alpha)$
4. Use the property:
$\displaystyle \tan(\beta - \alpha) = \dfrac{\tan(\beta) - \tan(\alpha)}{1+ \tan(\alpha) \cdot \tan(\beta)}$
5. Plug in the values you know. You should come out with: $\displaystyle (\beta - \alpha) \approx 52.125^\circ$
This formula belongs to a set of equations concerning the trigonometric functions of sums (or differences) of angles.
I learned it nearly 50 years ago - so I don't get it but I have ithow do you get it?
Fine!how does it work?
As I've mentioned you only have to plug in the known values:
$\displaystyle \tan(\beta - \alpha) = \dfrac{5 - \frac12}{1+ 5 \cdot \frac12}= \dfrac{\frac92}{\frac72}=\dfrac97
$
Now you can calculate the angle using the $\displaystyle \arctan-$ or $\displaystyle tan^{-1}$ function.
You don't have to remember the formula for $\displaystyle \tan(\beta-\alpha)$, provided that you know the formulas for $\displaystyle \sin(\beta-\alpha)$ and $\displaystyle \cos(\beta-\alpha)$ (which are definitely things that are useful to remember). In fact $\displaystyle \tan(\beta-\alpha) = \frac{\sin(\beta-\alpha)}{\cos(\beta-\alpha)} = \frac {\sin\beta\cos\alpha-\cos\beta\sin\alpha} {\cos\beta\cos\alpha+\sin\beta\sin\alpha}$. Now divide top and bottom by $\displaystyle \cos\beta\cos\alpha$ and you get the result $\displaystyle \tan(\beta-\alpha) = \frac{\tan\beta-\tan\alpha}{1+\tan\beta\tan\alpha}$.