.
One of the parallel sides of an isosceles trapezoid inscribed in circle
intercepts an arc with a measure of 80 degrees, while the other side is a diameter.
Find the measure of the angle formed by the diagonal of the trapezoid.
~~~~~~~~~~~~~~~~~~
Make a sketch.
Let the trapezoid be ABCD, with the diameter of the circle AB as the longer base and
the chord CD as the shorter base.
Let P be the intersection point of diagonals AC and BD of the trapezoid.
They want you find the angle APB.
Notice that the arcs AD and BC each has the measure of = 50 degrees,
since the trapezoid is isosceles.
So, the angles BAC and ABC have measures half of 50 degrees each, i.e. 25 degrees,
since they are inscribed angles leaning the arcs of 50 degrees.
Thus the triangle APB has the base angles PAB and PBA of 25 degrees each.
It implies that the angle APB is (180-25-25) = 130 degrees.
So, the problem is just solved.
The obtuse angle between the diagonals of the trapezoid is 130 degrees;
the corresponding acute angle is 180-130 = 50 degrees.
Solved.
///////////////
From the context, the meaning of the problem is MORE THAN OBVIOUS, so you, the reader,
feel free to use the solution in my post, disregarding critical notices by @greenestamps.
By the way and in addition, the meaning of the problem is CLEARLY DECIPHERED in my solution,
so any doubts are excessive . . .