The purpose of this question was to show the power of using exterior angles: It can make questions like these very simple! The method Professor Loh shows later in the video, using interior angles, is a little more complicated, but it's still worth knowing how to make use of interior angles.
The first step is to know what is the sum of all the interior angles in a polygon with \(n\) sides. You can try to look for a pattern:
For a triangle, the sum of angles is \(180.\)
For a quadrilateral, the sum is \(360\) (think of a square).
<Pentagons may be difficult, skip for now.>
For a hexagon, the sum is \(720\) (think of a regular hexagon!).
Based on this pattern, you can guess that the sum of angles goes up by \(180\) every time you add a side! But why? The clever reason comes from dividing the polygon into triangles.
You can see Professor Loh do this at 2:18. Every time you add another side, you get another triangle that makes up the polygon. And, adding up the angles of the polygon is the same as adding up the angles of every triangle in the "triangulation" (make sure this step makes sense!).
That means that the sum of the angles is just (sum of angles inside a triangle) \( \times \) (number of triangles in the polygon). Since the angle sum for a triangle is \(180,\) it's just \(180 \times \text{ number of triangles}.\)
The number of triangles that an \(n\)-sided polygon is divided into is just \((n-2)\). You can figure this out by looking at a pattern (\(n=3\) is \(1\) triangle, \(n=4\) is \(2\) triangles, etc.). Then the formula for the sum of angles is just \(180 \times (n-2).\) This is a very important formula! Make sure you understand why it works instead of memorizing it.
Back to the problem in the video: We don't know how many sides there are, so let's say there are \( n\) sides. Then since every angle is \(170,\) the sum of angles is \(170n.\) But we just derived a formula for the sum of angles: It's \(180(n-2).\) So we can set these expressions equal to each other:
$$ 170n = 180(n-2) $$
Now we just solve this equation and we are done!
I hope that helped. Happy learning!