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.
<pre><font size = 4 color = "indigo"><b>
{{{drawing(300,300,-2,13,-2,13, line(0,0,7,0),line(7,0,12,8),line(5,8,12,8),
line(0,0,5,8),graph(300,300,-2,13,-2,13),
locate(.5,1,A), locate(6.5,1,B), locate(12,9,C), locate(5,9,D)

 )}}} 

First draw a perpendicular DE to AB

{{{drawing(300,300,-2,13,-2,13, line(0,0,7,0),line(7,0,12,8),line(5,8,12,8),
line(0,0,5,8),graph(300,300,-2,13,-2,13),
locate(.5,1,A), locate(6.5,1,B), locate(12,9,C), locate(5,9,D),
line(5,8,5,0), locate(4.3,1,E)

 )}}} 

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 {{{TANGENT = (OPPOSITE)/(ADJACENT)}}}

{{{tan(DAB) = (DE)/(AE) = 8/5 = 1.6}}}

Now, with calculator in DEGREE mode, use the {{{TAN^(-1)}}}
key and get 57.99461679° which to the nearest degree is 58°

Edwin</pre>