In an isosceles trapezoid ABCD, side AB is equal to the midline of the trapezoid, AD = 3BC. Find angle A.

Let the smaller base of the BC of the trapezoid ABCD have a length of X cm, then, by condition, AD = 3 * X cm.

Since KM is the middle line of the trapezoid, KM = (BP + BC) / 2 = (3 * X + X) / 2 = 2 * X.

By condition, the lateral side of the trapezoid is equal to the midline, AB = KM = CD.

From point B we lower the height of the trapezoid to the base of AD.

AB = CD = KM.

Since the trapezoid is isosceles, then AH = (AD – BC) / 2 = (3 * X – X) / 2 = X.

Consider a triangle ABH, in which the hypotenuse AB = 2 * X, and the leg AH = X.

Cos A = BH / AB = X / 2 * X = 1/2.

Angle A = 60.

Answer: Angle A = 60.