Reduce the equation

Given equation is $x+\sqrt{3} y-4=0$

If the equation is in the form of ax + by = c, to get into the normal form, we should divide it by $\sqrt{a^{2}+b^{2}}$ so now

Divide by

$\sqrt{\sqrt{3}^{2}+1^{2}}=2$

Now we get

$\Rightarrow \frac{x}{2}+\frac{\sqrt{3} y}{2}=1$

This is in the form of

$x \cos \alpha+y \sin \alpha=p$

Where

$\cos \alpha=\frac{1}{2} \Rightarrow \alpha=\frac{\pi}{3}$

And p = 1

Conclusion: $\alpha=\frac{\pi}{3}$ and $p=1$