Question 143302: The vertices of parallelogram ABCD are A(0,0), B(7,0), (12,8), and D(5,8). Find, to the nearest degree the measure of angle DAB.
I don't get this at all! All I know is that the measure of side AB is 7 and the measure of side CD is also 7.
Answer by Edwin McCravy(20056)   (Show Source):
You can put this solution on YOUR website! The vertices of parallelogram ABCD are
A(0,0), B(7,0), (12,8), and D(5,8). Find,
to the nearest degree the measure of angle DAB.
I don't get this at all! All I know is that the
measure of side AB is 7 and the measure of side CD is also 7.
First draw a perpendicular DE to AB
Because D's coordinates are (5,8), we know that AE=5, DE=8
ADE is a right triangle.
DE is the side opposite angle DAB and
AE is the side adjacent angle DAB,
and since 
Now, with calculator in DEGREE mode, use the 
key and get 57.99461679° which to the nearest degree is 58°
Edwin
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