Find the lateral side AB of the trapezoid ABCD, if the angles ABC and BCD are 150

Find the lateral side AB of the trapezoid ABCD, if the angles ABC and BCD are 150 degrees and 45 degrees, respectively, and CD = 12 √2.

1. Draw the heights BP and DQ of the trapezoid from vertices B and D.

2. In a right-angled triangle DQC, the leg DQ is equal to the product of the hypotenuse CD and the sine of the opposite angle ∠C:

DQ = CD * sin∠C = 12√2 * sin (45 °) = 12√2 * √2 / 2 = 12.

3. The heights BP and DQ of the trapezoid drawn to the parallel sides are equal:

BP = DQ = 12.

4. The sum of the one-sided angles ABC and BAP is 180 °:

∠ABC + ∠BAP = 180 °;

∠BAP = 180 ° – ∠ABC = 180 ° – 150 ° = 30 °.

5. In a right-angled triangle ABP, the leg BP, opposite 30 °, is equal to half the hypotenuse:

BP = 1/2 * AB;

AB = 2 * BP = 2 * 12 = 24.

Answer: 24.