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?

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.
