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 °.