Answer
First five terms: $\quad 60^{\circ},\ 90^{\circ},\ 108^{\circ},\ 120^{\circ},\ 128.57^{\circ}$
$T_{n}=180(n-2)$
$T_{12}=1800^{\circ}$
Work Step by Step
The first regular polygon is an equilateral triangle (n=3):
$a_{3}=\displaystyle \frac{180(3-2)}{3}=60^{\circ}$
$a_{4}=\displaystyle \frac{180(4-2)}{4}=90^{\circ}$
$a_{5}=\displaystyle \frac{180(5-2)}{5}=108^{\circ}$
$a_{6}=\displaystyle \frac{180(6-2)}{6}=120^{\circ}$
$a_{7}=\displaystyle \frac{180(7-2)}{7}=(\frac{900}{7})^{\circ}\approx 128.57^{\circ}$
The total measure of interior angles is:
(number of vertices)$\times$(measure of an interior angle)
$T_{n}=a_{n}\displaystyle \cdot n=\frac{180n(n-2)}{n}$
$T_{n}=180(n-2)$
So for n=12:
$T_{12}=180(12-2)=1800^{\circ}$
