Find angles in trapezoid with sides' lengths equal $15$, $7$, $7$, $8$ 
What are the interior angles in the trapezoid with sides whose lengths equal  $15$, $7$, $7$, $8$ (sides with lengths $15$ and $7$ are parallel)?

I found this problem in elementary school's problem book. We should not use any trigonometry. Is there some simple solution, or the problem has some typo?
 A:  
Hint:
Choose $E\in [AB]$ such that $[AE]=8$ and $[EB]=7$. Now you have a rhombus with side length $7$ and an 8-8-7 isosceles triangle. From now on, I think you'll need trigonometry...
A: If we slide the two slant sides together, reducing the two parallel bases by equal amounts, we end up with a triangle with sides $7$, $8$, and $8$ - the two slanted sides and the difference between the parallel sides.
The angles in this isosceles triangle are $\arccos\left(\frac{3.5}{8}\right)\approx 64^\circ$, $\arccos\left(\frac{3.5}{8}\right)\approx 64^\circ$, and $2\arcsin\left(\frac{3.5}{8}\right)\approx 52^\circ$. Going back to the original trapezoid, the angles on the long base are the $52^\circ$ and one of the $64^\circ$ angles, leaving the supplementary angles $128^\circ$ and $116^\circ$ on the short base
Yeah, we're not getting those without trigonometry.
On the other hand, that same method suggests a fix - if that reduced triangle had sides $8,8,8$, we could just read off the $60^\circ$ angles of an equilateral triangle. That would require changing the length $7$ slant side into a length $8$ side. It's quite likely that this mistake was made.
A: Draw perpendiculars from the upper base of length 7 to the lower base of length 15. 
You cannot possibly find the interior angles without trigonometry.
