Find the angle B of an isosceles trapezoid ABCD with bases BC and AD if the angle D is 70 degrees.

In an isosceles trapezoid, the angles at the base are equal, that is, at the base AD, angle A = angle D = 70 degrees, at the base BC, angle B = angle C.
The sum of all interior angles of a quadrilateral is 360 degrees. Then:
angle A + angle B + angle C + angle D = 360 degrees.
Let’s denote angle A and angle B as x. Then:
70 + x + x + 70 = 360;
2x = 360 – 140;
2x = 220;
x = 220/2;
x = 110.
Angle B = angle C = x = 110 degrees.
Answer: angle B = 110 degrees.