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.
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 180o multiplied by the number of triangles. We must subtract the 360o 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 180o, we multiply by the number of vertices or sides. Since we are measuring the interior angles we must subtract the 360o, 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 180o multiplied by the number of triangles, we must then subtract the 180o added by creating an additional vertex on a line of the octagon.
The third formula is:
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 180o 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 180o. 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 180o, 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 180o, which results in a sum greater than 180o 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 180o.
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 180o. The method do not seem workable in the second octagon because there are 2 interior angles greater than 180o. There are no one place for an additional vertex that will make the method work.
Applying the third formula:
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 180o. 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 180o, we can accommodate as following:
Follow the shape clockwise starting at R1 and ending at Q1. When back at Q1 we are facing the same way as at R1, which mean we turned a full 360o and the exterior angles must add up to 360o. 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 U1 we are already facing the same direction as we did when starting at R1, which mean we have already turned 360o. At vertex V1 and W1 we turn to the opposite side (right instead of left), turning backwards in our 360o 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 V1 and W1 where the 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 180o and is over-lapped by the exterior angle. If we view the exterior angles at V1 and W1 as negative because they are inside the shape and are turning against the general direction, and therefore subtract it from the greater than 1800 interior angle, the angle pair's sum end up adding up to 180o and we are left with 8 angle pair with each the sum of 180o. Since we are measuring interior angle, we now subtract 360o, the total sum of the exterior angles in the angle pairs.
180*n - 360, DOES NOW APPLY TO ALL POLYGONS