Question

найдите угол АВС равнобедренной трапеции АВСD,если диагональ АС образует с основанием АD и боковой стороной СD углы равные 20 градусов и 100 градусов соответственно

Answer 1

Угол САД= 20 градусов
угол АСД=100 градусов

Угол АДС = 180 - 100-20= 60 градусов

 У равнобокой трапеции углы приосновании равны , угол ВАД= углу СДА =60 градусов

 Угол ВСА = углу ВАС =60-20= 40 градусов это накрест лежащие углы

 В треугольнике АВС ,угол ВАС= 40 , угол ВСА=20 , угол АВС= 180- 40-20= 120 градусов

Отвнт 120 градусов

Answer 2

Finding the angle ABC of an isosceles trapezoid ABCD

To find the angle ABC of the isosceles trapezoid ABCD, we are given that the diagonal AC forms angles of 20 degrees and 100 degrees with the base AD and the side CD, respectively.

Let's analyze the given information and find the angle ABC step by step.

1. We know that the sum of the angles in a triangle is 180 degrees. Therefore, the sum of angles ACD and CDA is 180 degrees. 2. We are given that angle ACD is 20 degrees. So, angle CDA is 180 degrees - 20 degrees = 160 degrees. 3. In an isosceles trapezoid, the base angles (angles BAC and BCA) are congruent. 4. We are given that angle CDA is 160 degrees. Therefore, angle BCA is also 160 degrees. 5. The sum of angles BCA and ABC is 180 degrees, as they form a linear pair. 6. Therefore, angle ABC is 180 degrees - 160 degrees = 20 degrees.

Hence, the angle ABC of the isosceles trapezoid ABCD is 20 degrees.

Please let me know if you need any further assistance!