In a trapezoid ABCD with bases BC and AD, it is known that ∠A = 60 ◦, ∠D = 30 ◦, AD = 18, BC = 4. Find AB.

Let’s build the heights of the BM and CH of the trapezoid.

Quadrangle MBCH is a rectangle, then MH = BC = 4 cm.

Then AM + DH = AD – MH = 18 – 4 = 14 cm.

Let AM = X cm, then DH = (14 – X) cm.

In right-angled triangles ABM and CDH, we express the heights of BM and CH.

Let BM = CH = h see.

tg60 = h / AM = h / X.

tg30 = h / DH = h / (14 – X).

h = X * tg60.

h = (14 – X) * tg30.

X * tg60 = (14 – X) * tg30.

X * √3 = (14 – X) / √3.

3 * X = 14 – X.

4 * X = 14.

X = AM = 14/4 = 7/2 cm.

Cos60 = AM / AB.

AB = AM / Cos60 = (7/2) / (1/2) = 7 cm.

Answer: The length of the segment AB is 7 cm.