Author:
Tina Lyder

To show how application of certain formulas can be used to measure interior angles of all polygons.

Images are used as examples to show how to use the formulas 180*n - 360, (n-1)*180 - 180, and (n-2)*180 to measure interior angles of regular and irregular polygons.

Tutorial

This packet is using octagons as examples to demonstrate how to measure interior angles of regular convex polygons with three different formulas, 180*n - 360, (n-1)*180 - 180, and (n-2)*180. The application of these formulas involve methods of dividing the shapes into triangles. When trying to apply these methods to irregular polygons like concave octagons we are running into problems, and further examples explore and describe these problems, and show how to modify two of the techniques to accommodate for these problems and so make applicable to all polygons.

Measuring interior angles of a convex polygon can be described as one of the following formulas, 180*n - 360, (n-1)*180 - 180, or (n-2)*180:

180*n - 360

By placing an additional vertex in the octagon we creates as many triangles as there are sides to the octagon. The sum of the interior angles of each created triangle is 180^{o} multiplied by the number of triangles. We must subtract the 360^{o} added by creating an additional vertex in the octagon.

The same formula, 180*n - 360, will be explained by the following:

We can see each angle as a pair of an interior and an exterior angle. The sum of each vertex's angle pair is 180^{o}, we multiply by the number of vertices or sides. Since we are measuring the interior angles we must subtract the 360^{o}, which is the total sum of the exterior angles.

The second formula is:

(n-1)*180 - 180

By placing an additional vertex on the line of the octagon we create we create one less triangle than there are sides. The sum of the interior angles of each created triangle is 180^{o} multiplied by the number of triangles, we must then subtract the 180^{o} added by creating an additional vertex on a line of the octagon.

The third formula is:

(n-2)*180

We divide the octagon into triangles using an already existing vertex, creating 2 less triangles than there are sides. The sum of the interior angles of each created triangle is 180^{o} multiplied by the number of triangles, since we are not adding any additional vertex in this method there is nothing to subtract.

The formulas will apply to any regular polygon, but will the formulas and diagrams also work for an irregular polygon as a concave polygon? A concave polygon has at least one interior angle greater than 180^{o}. Let us try to apply the first formula to the two following concave octagons.

180*n - 360

In the first shape we can make the formula work by placing a new vertex strategically (I), but we can not place the additional vertex anywhere in the octagon. If our additional vertex is placed as (R) the method will not work. The second shape has more than one interior angle greater than 180^{o}, and it will not be possible to place a vertex strategically to make the method work.

We apply the same formula, 180*n - 360, to the concave octagons using the method with angle pairs:

When looking for the 8 angle pairs in the first concave octagon, one of the interior angles (H), seems to be found on the inside of the octagon. Furthermore, the interior angle (H) is greater than 180^{o}, which results in a sum greater than 180^{o} for this angle pair. In the second shape we have the same problem as in the first shape, but here we have 2 vertices (V and W), which seem to have angle pairs where the sum of the angle pairs exceed 180^{o}.

Applying the formula:

(n-1)*180 - 180

In the first octagon we can place our additional vertex strategically as (I) and make the formula work, but we can not place the additional vertex on any line in the octagon. If we place it as (R) the method breaks down because of the interior angle greater than 180^{o}. The method do not seem workable in the second octagon because there are 2 interior angles greater than 180^{o}. There are no one place for an additional vertex that will make the method work.

Applying the third formula:

(n-2)*180

Working with the third formula, we get the same result as earlier. Here we are using an already existing vertex, and we can strategically choose a vertex (H) that will make the methods workable, but not any one vertex can be used, for example (G). In the second octagon it is not possible to divide the shape into triangles using one existing vertex.

Let us modify two of the above methods so they will apply to all polygons.

First the formula, (n-2)*180:

When we investigated the method earlier, we divided the shape into triangles with lines from one already existing vertex, and found that there was not one vertex in this octagon from where we could divide the octagon because of one or more interior angles greater than 180^{o}. Instead of dividing the shape from one single vertex we connect all the already existing vertices, H is already connected to A, A to G, G to B, B to F, F to C, C to E, E is already connected to D. We have created 6 triangles.

(n-2)*180, DOES NOW APPLY TO ALL POLYGONS

If we consider the formula: 180*n - 360, and apply the method, viewing the angle of each vertex as an angle pair with the sum of 180^{o}, we can accommodate as following:

Follow the shape clockwise starting at R_{1} and ending at Q_{1}. When back at Q_{1} we are facing the same way as at R_{1}, which mean we turned a full 360^{o} and the exterior angles must add up to 360^{o}. The exterior angle in each angle pair is marked with red shading, and signifies the degree turned to follow the next sides of the shape. Note that when turning at vertex U_{1} we are already facing the same direction as we did when starting at R_{1}, which mean we have already turned 360^{o}. At vertex V_{1} and W_{1} we turn_{ }to the opposite side (right instead of left), turning backwards in our 360^{o} turn. It is like going in the opposite direction on the number-line and can be thought of as a negative turn. It is seen clearly at V_{1} and W_{1} where t_{}he exterior angle (the degree turned) is inside of the shape. Furthermore, the interior angles, which are green shaded, at these two vertices are greater than 180^{o} and is over-lapped by the exterior angle. If we view the exterior angles at V_{1} and W_{1} as negative because they are inside the shape and are turning against the general direction, and therefore subtract it from the greater than 180^{0} interior angle, the angle pair's sum end up adding up to 180^{o}^{ }and we are left with 8 angle pair with each the sum of 180^{o}. Since we are measuring interior angle, we now subtract 360^{o}, the total sum of the exterior angles in the angle pairs.

180*n - 360, DOES NOW APPLY TO ALL POLYGONS

^{}