posted by jeff .
the bases of a trapezoid are 22 and 12 respectively. The angles at the extremities of one base are 65^ and 40^ respectively. Find the two legs
Since the long base is 10 more than the short one, if you eliminate the interior rectangle of side 12, you have two triangles whose bases add up to 10.
You know two of the angles, so the 3rd angle is 75°. Now just use the law of sines to find the other two sides (the legs of the trapezoid):
10/sin75° = a/sin65° = b/sin40°
Let the trapezoid ABCD be of height h.
AB = 22
CD = 12
DAB = 65°
ABC = 40°
Construct a diagram and consider the trigonometry:
(1) AB = CD + h cot(DAB) + h cot(ABC)
(2) BC = h sec(ABC)
(3) DA = h sec(DAB)
Substitute the values into (1) to solve for h.
Substitute h into (2) and (3) to find the lengths of the other two sides.