$(360-(\frac{360}{n}+180))n=(n-2)180$

Question

Programmers and engineers point of view.

I'm a programmer and just made a formula to compute a shapes total internal angles using n = number of angles of shape: $(360-(\frac{360}{n}+180))n$.

I arrived with my formula by assuming all point of shapes are derived from a combination of 360=180+externalAngle+internalAngles where the sum of externalAngles = 360

An engineer suggested: $180(n-2)$

How do I prove they are same? Preferred solution in algebra, if possible.

Answer 1

$(360-(\frac{360}{n}+180))n=360n-360-180n=180n-360=180(n-2)$

Answer 2

$(360-(\frac{360}{n}+180))n = (360-\frac{360}{n}-180)n = (180-\frac{360}{n})n = 180(1-\frac{2}{n})n = 180\frac{n-2}{n}n = $

$$180(n-2)$$