How do you find m in terms of n, supposing that angle CEF is the interior angle of another regular polygon with m sides?

Question

angle CEF = $\frac{720 d e g r e e s}{n}$ $(720 degrees)/n$ as a function of n

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Answer 1 · 2018-04-30T14:46:01

Total of $m$ $m$ interior angles of a polygon of m sides will be $(2 m - 4) \cdot 90^{\circ}$ $(2m-4)*90^@$ .

Given that angle CEF is the interior angle of a regular polygon with m sides.

We can write

$∠ C E F = \frac{(2 m - 4) \cdot 90^{\circ}}{m}$ $angleCEF=[(2m-4)*90^@]/m$

Again it is given $∠ C E F = \frac{720^{\circ}}{n}$ $angleCEF=720^@/n$

Hence

$∠ C E F = \frac{720^{\circ}}{n} = \frac{(2 m - 4) \cdot 90^{\circ}}{m}$ $angleCEF=720^@/n=[(2m-4)*90^@]/m$

$\Rightarrow \frac{720^{\circ}}{n} = \frac{(2 m - 4) \cdot 90^{\circ}}{m}$ $=>720^@/n=[(2m-4)*90^@]/m$

$\Rightarrow \frac{8}{n} = \frac{2 m - 4}{m}$ $=>8/n=(2m-4)/m$

$\Rightarrow \frac{8}{n} = 2 - \frac{4}{m}$ $=>8/n=2-4/m$

$\Rightarrow \frac{4}{m} = 2 - \frac{8}{n}$ $=>4/m=2-8/n$

$\Rightarrow \frac{2}{m} = 1 - \frac{4}{n}$ $=>2/m=1-4/n$

$\Rightarrow \frac{2}{m} = \frac{n - 4}{n}$ $=>2/m=(n-4)/n$

$\Rightarrow m = \frac{2 n}{n - 4}$ $=>m=(2n)/(n-4)$