Getting an angle from a non-regular pentagon given two angles and the length of all the sides Is it possible to solve for angle x given angle A, a right angle, and the length of all the sides?

*note that the legnth of all the sides are known, however no side has the same length
 A: Yes, you can construct a pentagon with  the given information.
Let us label the vertices in the given graph, starting with $A$ and going counter clockwise, so that the right angle is $B$ and then $C$, $D$, $E$.
We can easily locate the vertices $A,B,C$ and $E$ given the sides and the angles at $A$ and $B$
The only remaining vertex is $D$ and we can locate that by drawing two circles centered at $C$ and $E$ with radii equal to  the given sides and find the point of intersection. 
Once you constructed your pentagon, you can measure the angle or you can solve for it mathematically using laws of sine and cosine.
A: Let the pentagon be $ABCDE$ with $A$ at the angle labeled $A$ and $E$ at the given right angle.
First draw diagonal $\overline {AD}$ completing right $\triangle ADE$.  Thereby obtain $AD$ from the Pythagorean Theorem and measure $\angle EAD$ by the usual trigonometric laws.  Subtract that angular measure from the full measure of $\angle EAB$ to measure $\angle DAB$.
That reduces your problem to a quadrilateral with four known sides and one known angle ($\angle DAB$).  Draw diagonal $\overline{BD}$ and get its length from the Law of Cosines on $\triangle ABD$.  You now know all the sides of that triangle and also $\triangle BCD$, so you solve for their angles and add the two results at $B$ to get $x$.
