anglegeometrypolygonsquadrilateral
I have the measurements of the four sides of an irregular polygon and I need to find out the size of each interior angle.
I know the sum of the angles is 360 degrees but because it's not a regular polygon I don't know how to calculate each angle.
The measurements of each side are 882.9, 80, 576.8 and 293.3.
However, it is confirmed that one of the angles is 90 degrees.
How do I find the other 3 angles?
Please refer to the image below for reference.
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.
At each vertex of the polygon, the interior angle and exterior angle must sum to $180$. You can use this to find the value of each exterior angle.
Best Answer
There is no such quadrilateral. For draw the hypotenuse of the right triangle with legs $882.9$ and $80$. This hypotenuse has length $l$ greater than $882.9$.
But $293.3+576.8\lt 882.9\lt l$, contradicting the triangle inequality: the sum of any two sides of a triangle is greater than the third side.