Find the angles of an isosceles trapezoid if its side is 7 cm, and a diagonal of 7√3 cm makes an angle of 30 degrees with the base.

An isosceles trapezoid ABCD is given.
AB = CD = 7 cm – sides of the trapezoid.
BD = 7 * √3 – trapezoid diagonal.
Consider a triangle ABD. In it, we know the lengths of the two sides and the angle between them: BDA = 30 °.
We use the theorem of sines and find out what the angle BAD is equal to:
7 / sin 30 = 7 * √3 / sin (ABD);
sin (ABD) = 7 * √3 * sin 30/7 = √3 / 2;
ABD = 60 °.
Let us find the angle ABC knowing that in the sum with ABD they are 180 °, since they are adjacent to the side of the trapezoid.
ABC = 180 – ABD = 180 – 60 = 120 °
Answer: the angle at the lower base of the trapezoid is 60 °, the angle at the upper base is 120 °.