Suppose someone starts at a point A, and walks towards the North direction for some distance $d$. Then turns right $\theta$ degrees, and repeats the process. How many turns will he make until he reaches the starting point?
From what I've seen in problems, you calculate the external angle of turning i.e $180-\theta$. Then calculate $n$ for $$(180-\theta)n \equiv 0 \mod 360$$
However I can't quite make sense of this in my head. Please help me understand the intuition and proof behind it. Thanks.
Suppose you have a regular polygon with interior angle $\theta$ and $n$ sides. The relation $$\frac{180(n-2)}{n}=\theta$$ holds. Multiplying by $n$ and distributing the 180 yields $$180n-360=\theta n.$$ Rearranging you obtain $$180n-\theta n=360.$$ Pulling out the $n$ you get $$(180-\theta)n=360.$$ Hence, if $\theta$ is your interior angle, then solving for $n$ in $$(180-\theta)n=0\mod{360}$$ will give you the value of $n$. If you want to work instead with the exterior angle $\theta_{\text{ext}}$ at which you turn then you need to see that $\theta=180-\theta_{\text{ext}}$. Substituting this into the equation above, we get $$(180-(180-\theta_{\text{ext}}))n=0\mod{360},$$ or $$\theta_{\text{ext}}n=0\mod{360}.$$ Hence, your trouble seems to be that you are getting the internal angle $\theta$ and the exterior angle $\theta_{\text{ext}}$ confused.