Thursday

July 24, 2014

July 24, 2014

Posted by **jeff** on Sunday, September 1, 2013 at 11:46pm.

- trigonometry -
**Steve**, Monday, September 2, 2013 at 4:06amSince 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°

- trigonometry -
**Graham**, Monday, September 2, 2013 at 4:23amLet the trapezoid ABCD be of height h.

Given:

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.

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