# Interior angles of irregular quadrilateral with 1 known angle

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.

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.
• Let us label the vertices. Bottom right, at the right angle, $A$. Top right, $B$. Extreme left $C$. And the last one, close to $A$ but $80$ to the left of it, $D$. Join the points $B$ and $D$ by a straight line. This line $BD$ is the hypotenuse of a right-angled triangle, so it has length that I called $l$ in the post, where $l\gt 882.9$. We could compute $l$ using the Pythagorean Theorem, but don't need to. (Continued) – André Nicolas Feb 26 '16 at 8:02
• (Continued) Now look at "triangle" $BCD$. I put triangle in quotes because it is an impossible triangle. For the sum of two of its supposed sides, $BC$ and $CD$, is less than the third side $BD$. Thus there is no such thing as a quadrilateral with sides as given and right angle at $A$. – André Nicolas Feb 26 '16 at 8:06