Angles in polygons in part two

Trigonometry Angles in polygons in part two

We know that if we add all of the external angles of any polygon together then we get $360^\circ$ .

There is a simple way to work out what we get if we were to add all of the internal angles of any polygon.

The dotted lines create triangles in the pentagons above. There are three triangles in both pentagons. We know that the internal angles in triangles add up to $180^circ$ and so the internal angles in a pentagon add up to $180^\circ \times 3 = 540^\circ$ . There are 5 internal angles in a pentagon and in a regular pentagon the angles are all the same, so in a regular pentagon, the internal angles are equal to $\frac{540^\circ}{5} = 108^\circ$ . This means that the external angle of a regular pentagon is $180 - 108 = 72^\circ$ . These are the same answers we got from calculating the external angle first.

We can apply this technique to find the sum of the internal angles within any polygon and the size of the internal angle for any regular polygon. The number of triangles we can create within a polygon is $the number of sides - 2$ . (Draw some polygons with different numbers of sides and draw triangles in them originating from a single vertex if you want to satisfy yourself that this is true.) If we let $n = the number of sides$ of a polygon then the sum of the internal angles of the polygon is $180(n - 2)^\circ$ . Dividing this number by $n$ gives the internal angle of a regular polygon.

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