Find the angle ABC of an isosceles trapezoid ABCD, if the diagonal AC makes

Find the angle ABC of an isosceles trapezoid ABCD, if the diagonal AC makes angles equal to 30 and 80 degrees with the base AD and the lateral side CD.

Let’s consider different solutions.
Option 1.
In the triangle ACD, two angles are known by condition, we find the third:
∠ D = 180 ° – (∠ CAD + ∠ CDA) = 180 ° – (30 ° + 80 °) = 70 °.
The trapezoid is isosceles by condition, ∠ D = ∠ A = 70 °.
The angles adjacent to one side add up to 180 °, find the angle B:
∠ B = 180 ° – ∠ A = 110 °.
Option 2.
In the triangle ACD, two angles are known by condition, we find the third:
∠ D = 180 ° – (∠ CAD + ∠ CDA) = 180 ° – (30 ° + 80 °) = 70 °.
BC || AD, AC – secant.
∠CAD = ∠ BCA = 30 ° (lying crosswise).
∠ С = ∠ BCA + ∠ CDA = 30 ° + 80 ° = 110 °.
The trapezoid is isosceles by condition, ∠ С = ∠ В = 110 °.
Answer: the angle ABC is 110 °.



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