The sum of the two angles of an isosceles trapezoid is 236 degrees. Find the smaller angle of the trapezoid

Let an isosceles trapezoid ABCD with bases AD and BC be given. 1. Since the bases of the trapezoid are parallel, then ∠A + ∠B = 180 °, since they are one-sided angles formed at the intersection of two parallel lines AD and BC secant AB. And since ∠B = ∠C, then ∠A + ∠C = 180 °. Thus: ∠A + ∠D = 236 °. Since ∠A = ∠D, we denote them as x: x + x = 236 °; 2 * x = 236 °; x = 236 ° / 2; x = 118 °. Then: ∠A = ∠D = x = 118 °. 1. Find the degree measure ∠B: ∠A + ∠B = 180 °; 118 ° + ∠B = 180 °; ∠B = 180 ° – 118 °; ∠B = 62 °. 1. ∠A… ∠B; 118 °> 62 °, so ∠B = ∠C is a smaller angle. Answer: ∠B = ∠C = 62 °.