# How to determine the side lengths of an irregular polygon when all interior angles are known?

1. Given an irregular polygon where all of the angles are known, how many side lengths need to be known, at minimum, to determine the length of the remaining sides?

2. Given all the angles and the requisite number of side lengths, how to actually calculate the remaining side length?

Example: a 5 Sided polygon's interior angles will add up to 540 degrees. ((5-2)*180=540).

Given the following interior angles:

AB 140 degrees
BC 144 degrees
CD 78 degrees
DE 102 degrees
EA 76 degrees


And knowing that Side A is 12 units long, can we determine the remaining side lengths? Or are more side lengths needed?

UPDATE:

Since you need three consecutive side lengths of a five sided figure, I'm adding three sides here so I can see an example of how the calculations are done for the remaining two sides:

Side A = 27 7/8"
Side B = 7"
Side c = 13 1/4"


For a $$n$$ sided polygon, you need all the angles in order and $$n-2$$ consecutive side lengths in order to construct the polygon. So, you need the lengths of sides $$B,C$$ or $$E,B$$ or $$D,E$$ to construct your polygon.

The best way to find out the length of the remaining side is by drawing diagonals and applying triangle laws (sine or cosine rule).

Consider the (very badly drawn) pentagon. It is not drawn to scale, but you get the idea. Here are the steps you will take to find out the lengths of $$D,E$$.

1.Find out length of $$X$$ using cosine rule in $$\Delta ABX$$.

2.Knowing $$\angle a,X,A,B$$, find out $$\angle e,\angle b$$ using sine rule.

3.$$\angle c = \angle\mbox{(between B,C)} -\angle b$$. So, $$\angle c$$ is known.

4.Repeat the whole procedure for $$\Delta CXY$$. Find out $$Y,\angle d, \angle f$$.

5.$$\angle g, \angle h$$ are easily calculated now.

6.$$\angle i$$ is known. Apply sine rule in $$\Delta DEY$$ to find out $$D,E$$-the two unknown sides. • Could you provide more details or a link on "drawing diagonals and aplying triangle laws"? I'm not familiar with how that works.
– Nick
Dec 11, 2012 at 18:33
• I've added an update with example dimensions of three consecutive sides.
– Nick
Dec 11, 2012 at 20:11
• I have added an explanation. Got it? Dec 11, 2012 at 20:48
• Got it! Thanks for the picture and details. +1 and accepted.
– Nick
Dec 11, 2012 at 21:44
• @dexter04 What about if we know there is an axis of symmetry? For example let's say there's a hexagon which can be folded in half horizontally to give overlapping halves, but not vertically. Only 2 unique angles are needed to determine the rest (there's 3 pairs of angles, so the 3rd pair can be determined from the other two), but what about the 4 unique lengths? By drawing two examples with same angles and one common side, I can see 1 side-length isn't enough, but it seems knowing 2 side-lengths can determine the remaining 4 side lengths. Is there a general rule/theorem/proof for all polygons? Nov 30, 2017 at 2:52

Here's a straightforward way of 'back-of-the-envelope'ing the number of sides needed: arbitrarily translate one of your polygon's points to the origin, and align its first side along the positive $x$ axis; it should be intuitively obvious (though it's a little trickier to prove) that this uses all the degrees of freedom that are available in the isometries of the plane. Once this has been done, then since we've set the 'root' of the figure to be the origin we have $(x_0, y_0) = (0, 0)$) and setting the first side along the $x$ axis has given us $y_1 = 0$; on the other hand, it should be clear (though again, a little tricky to prove) that all the other coordinates $\{(x_i, y_i): 1\leq i\leq n-1\}$ of the polygon are essentially unspecified. This means that there are $2(n-1)-1 = 2n-3$ degrees of freedom available to describe the polygon. (Alternately, you can think of this as the $2n$ degrees of freedom inherent in $n$ $2$-dimensional points minus the $3$ degrees of freedom in isometries of the plane.) Since there are $n-1$ degrees of freedom in the polygon's angles (the sum of the angles is constant, which removes one degree of freedom), then an additional $(2n-3)-(n-1) = n-2$ values — in other words, the length of all but $2$ sides — will need to be provided in order to eliminate all the degrees of freedom and completely specify the polygon.