In a quadrilateral, 2 opposite sides are parallel and are 57 and 55 mm.

Question

In a quadrilateral, 2 opposite sides are parallel and are 57 and 55 mm. One of the remaining side is 12 mm and makes 30 degrees to the horizontal, and another is 11 mm and makes 20 degrees to the horizontal. What is the angle made by parallel sides to the horizontal.

Answer 1

I am not 100% certain of this solution, but I think it is probably correct.

 

You need to make a drawing.  Call the left end of the longer base of the trapezoid A and the right end B.

Call the left end of the smaller base C and the right end D.

Extend the non-parallel sides of the trapezoid until they meet at a point E.

Draw the horizontal through A.

Extend line through points EDB until it meets the horizontal at F.

∠AEF = 130° (180° - 30° - 20°)

ΔABE is similar to ΔCDE.

Let y be the length of segment DE.

x/55 = (y+11)/57

Use the value for y and the law of sines to compute ∠ECD = ∠EAB (Remember ∠CDE = 130°)

The angle you need for this problem is ∠BAF and it = 30° - ∠EAB

 

There may be another way to solve this but I don't see it.

If you get another solution, especially one easier and/or better, I would certainly like to know about it.