Can Angles A and B In A Trapezium Be Solved Using Basic Geometry? 
Can angles A and B be solved? Neither the area nor the perimeter was given. Thank you very much if you can help! :)

If I'm not mistaken, I think the followings tell us that the angles $A,B$ are not determined. Set 
$$A(0,0),B(9,0),C(9+7\cos(180^\circ-B),7\sin(180^\circ-B)),D(4\cos A,4\sin A).$$
Then, we have
$$CD^2=5.9^2=\{9+7\cos(180^\circ-B)-4\cos A\}^2+\{7\sin(180^\circ-B)-4\sin A\}^2$$
$$\iff 5.9^2=9^2+7^2+4^2-126\cos B-72\cos A+56\cos A\cos B-56\sin B\sin A.$$
Here, even if we suppose that $0\le A,B\le 90^\circ,$ we have
$$5.9^2=9^2+7^2+4^2-126y-72x+56xy-56\sqrt{(1-x^2)(1-y^2)}$$
where $x=\cos A, y=\cos B, \sin A=\sqrt{1-x^2}, \sin B=\sqrt{1-y^2}.$
For example, if $A=90^\circ$, then $\cos B=y=39.19/70\approx 0.559857.$
If $A=45^\circ$, then $\cos B=y\approx 0.254432.$
Since we have at least two concrete examples, it's obvious that there are infinitely many pairs $(A,B)$ because $C$ is on the circle whose center is $B$ with radius $7$ and $D$ is on the circle whose center is $A$ with radius $4$. 
