In a rectangular trapezoid ABCD with bases AD and BC, angle A = angle B = 90 degrees

In a rectangular trapezoid ABCD with bases AD and BC, angle A = angle B = 90 degrees, angle ACD = 90 degrees, AD = 4cm, BC = 16 cm.Find angles C, D

From the top of the trapezoid, draw the height CH.

Since the quadrangle ABCН is a rectangle, then AH = BC = 4 cm.

Then DН = AD – AН = 16 – 4 = 12 cm.

In a rectangular, by condition, triangle of AВD, the height of the CH is drawn from the top of the right angle, then CH^2 = AH * DН = 4 * 12 = 48.

CH = 4 * √3 cm.

Then in a right-angled triangle CDН tgD = СН / DН = (4 * √3) / 12 = √3 / 3.

Angle ADС= arctan (√3 / 3) = 30.

Since the sum of the angles of the trapezoid at the lateral side is 1800, the angle of ВСD = 180 – ADС = 180 – 30 = 150.

Answer: Angle C = 150, Angle D = 30.