trigonometry
posted by joe .
the bases of a trapezoid are 22 and 12 respectively. The angles at the extremities of one base are 65 degree and 40 degree respectively find the two legs. Answer using law of sines pls

Construct the trapezoid ABCD ,where AB  CD
AB = 12 and CD = 22
angle C=65 and angle D = 40
Draw AE  BC where E is on CD
So now ABCE is a parallelogram, and CE = 12
which makes ED = 10
Now look at triangle AED, by corresponding angles
angle AED = 65°, angle D = 40 leaving angle DAE = 75°
by sine law:
AD/sin65 = 10/sin75
AD = 10sin65/sin75 = 9.38
by sine law:
AE/sin40 = 10/sin75
AE = 6.65
but BC = AE, (gram
So the side adjacent to the 65° angle is 6.65, the side adjacent to the 40° angle is 9.38
check my arithmetic, I am only on my first coffee. 
wow this really helps!! :D I have the same problem too haha thanks!!!

Hey thanks! I have the same problem. never thought it would be this easy.

tnx bro!

where'd you get that 10 and 12
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