In trapezoid ABCD, angle A: angle B: angle C≈1: 2: 3 find the angles of the trapezoid

Let the bases of the trapezoid be BC and AD.

Angles A, B and C are related as 1: 2: 3. Let the angle A be equal to x, then the angle B will be equal to 2x, and the angle C will be equal to 3x.

Since angles A and B are angles at the lateral side AB, their sum is 180 °.

x + 2x = 180; 3x = 180; x = 180: 3 = 60 ° (angle A).

Angle B is 2x = 2 * 60 = 120 °.

Angle C is equal to 3x = 3 * 60 = 180 ° (this cannot be, the angle of a trapezoid cannot be deployed).

Let the bases of the trapezoid be AB and CD.

Angles A, B and C are related as 1: 2: 3. Let angle A be equal to x, then angle B will be equal to 2x, and angle C will be equal to 3x.

Since angles B and C are angles at the side of the BC, their sum is 180 °.

2x + 3x = 180; 5x = 180; x = 180: 5 = 36 ° (angle A).

Angle B is 2x = 2 * 36 = 72 °.

Angle C is 3x = 3 * 36 = 108 °.

Angles A and D are the angles at the lateral side AD, their sum is 180 °.

Angle D = 180 – 36 = 144 °.

Answer: The angles of the trapezoid are 36 °, 72 °, 108 ° and 144 °.



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