Find the side AB of trapezoid ABCD if the angles ABC and BCD are 60 ° and 135 °, respectively, and CD = 24.

Let the trapezoid ABCD be given, which has bases AD and BC. From the condition of the problem, it is known that the angles ABC and BCD are equal to 60 ° and 135 °, respectively, and CD = 24. From the vertex of the angle BCD to the base of AD we draw the height SK, we obtain a right-angled triangle SKD, in which ∠КСD = ∠BCD – ∠BCК = 135 ° – 90 ° = 45 °. Then ∠СDК = 90 ° – ∠КСD = 90 ° – 45 ° = 45 °, since the sum of the acute angles in a right-angled triangle is 90 °. This means that the triangle is isosceles and the leg

CK = СD ∙ sin 45 ° = 24 ∙ 0.5 ∙ √2 = 12 ∙ √2.

From the vertex of the angle BAD to the base of the BC, draw the height AP = SK, we get a right-angled triangle ABP, in which the hypotenuse

AB = AP: sin 60 ° = (12 ∙ √2): (√3) / 2 = 8 ∙ √6 ≈ 19.6.

Answer: ≈ 19.6 is the length of the side AB.