Find the value of the angle at the greater base of the isosceles trapezium ABCD

Find the value of the angle at the greater base of the isosceles trapezium ABCD, the area of which is 15√3 cm2 wound. Given AB = CD, BC = 2cm AD = 8cm.

Let’s build the height BH of the trapezoid ABCD.

Since the trapezoid is isosceles, the BH height divides the base into two segments, the length of the smaller of which is equal to the half-difference of the base lengths.

AH = (AD – BC) / 2 = 6/2 = 3 cm.

The area of the trapezoid is: S = (BC + AD) * BH / 2.

ВН = 2 * S / (ВС + АD) = 2 * 15 * √3 / 10 = 3 * √3 cm.

In a right-angled triangle ABH, tgA = BH / AH = 3 * √3 / 3 = √3.

Angle A = arctan √3 = 60.

Answer: The angle for a larger base is 60.