Find the side AB of trapezoid ABCD if the angles ABC and BCD are 45 ° and 120 °, respectively, and CD = 34.
| November 19, 2020 | education
Triangle CMD cosMCD = h / CD
h = CD cosMCD = CD cos (beta-90)
Triangle ANB sin alpha = h / AB
AB = h / sin alpha = (CD cos (beta -90)) / sin alpha
Alpha = 45
Betta = 120
AB = (34 • cos (120-90)) / sin45 = 34 • (√3 / 2) / √2 / 2 = 34 √3 / √2
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