$\begin{align}  & \text{Sum of interior angle or a regular polygon is given as }(n-2){{180}^{\circ }} \\ & (n-2){{180}^{o}}={{1800}^{o}} \\ & n-2=10,\text{ }n=12 \\ & \text{Each interior angle will be }\frac{1800}{12}={{150}^{o}} \\\end{align}$Using the diagram aboveThe exterior angle will be 180⁰ – 150⁰ =30^o Alternatively,The sum of the exterior polygon is 360^oSince the polygon has 12 sides12 × θ  =360^o$\theta =\frac{{{360}^{o}}}{12}={{30}^{o}}$