In an isosceles trapezoid, the diagonal makes an angle of 120 degrees with the lateral side.

In an isosceles trapezoid, the diagonal makes an angle of 120 degrees with the lateral side. The lateral side is equal to the smaller base. Find the corners of the trapezoid.

Since, according to the condition, the lateral side is equal to the smaller base, then the ABC triangle is isosceles, and therefore the angle BAC = BCA.

The angle ACB is equal to the angle DAC as criss-crossing angles at the intersection of parallel straight lines BC and AD secant AC.

Then the angle BAC = DAC, which means AC is the bisector of the angle BAD.

Let the angle DAC = X0, then the angle BAD = ADC = 2 * X.

In triangle ACD, the sum of the interior angles is: 180 = (120 + X + 2 * X).

3 * X = 180 – 120 = 60.

X = 60/3 = 20.

Angle BAD = ADC = 2 * 20 = 40.

In a trapezoid, the sum of adjacent angles is 180, then the angle ABC = 180 – 40 = 140.

Answer: The angles of the trapezoid are 40, 140.



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