Trapezoid ABCD with base AB is inscribed in a circle. angle ADB = 65, angle DBC = 35, then angle A is …

After the diagonals of the trapezoid ABCD intersect, we have formed a triangle ODB, the two angles of which are known to us. These are the angles ODB and DBO. Their degree measures are 65 ° and 35 °, respectively.
Knowing them, we can find the third corner of DOB:
∠DОВ = 180 ° – ∠ODВ – ∠DВО.
∠DOV = 180 ° – 65 ° – 35 ° = 80 °.
Since the angles COA and DOB are formed by the intersection of two segments, they will have the same degree measure and it is equal to 80 °.
The AOB triangle is isosceles. The OAB angle will be 40 °.
Find angle A:
40 ° + 35 ° = 75 °.
ANSWER: ∠А = 75 °.