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

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

Let’s draw the height of the CH trapezoid.

The segment AH = CB = 4 cm, since the quadrilateral ABCN is a rectangle.

DН = АD – АН = 16 – 4 = 12 cm.

By the property of the height of a right-angled triangle drawn to the hypotenuse:

CH ^ 2 = AH * DH = 4 * 12 = 48.

CH = 4 * √3 cm.

Let’s define the tangent of the angle СДН.

tgСDН = СН / DH = 4 * √3 / 12 = √3 / 3

Angle СDН = 30.

Then the angle DСН = 90 – 30 = 60.

Angle DСВ = 90 + 60 = 150.

Answer: The angle DCB is 150, the angle CDA is 30.