Questions: Найдите угол ABC равнобедренной трапеции ABCD, если диагональ AC образует с основанием AD и боковой стороной CD углы, равные 20° и 100° соответственно.

Solution Steps

We are given an isosceles trapezoid \(ABCD\) with bases \(AD\) and \(BC\). The diagonal \(AC\) forms angles of \(20^\circ\) with base \(AD\) and \(100^\circ\) with side \(CD\).

In triangle \(ACD\), we know:

• \(\angle CAD = 20^\circ\)
• \(\angle ACD = 100^\circ\)

Since the sum of angles in a triangle is \(180^\circ\), we can find \(\angle DAC\): \[ \angle DAC = 180^\circ - 20^\circ - 100^\circ = 60^\circ \]

Since \(ABCD\) is an isosceles trapezoid, the angles at the bases are equal. Therefore, \(\angle ABC = \angle DAB\).

From the previous step, we know: \[ \angle DAB = 60^\circ \]

Final Answer