Question 919189
{{{drawing(400,2400/7,-7,7,-2,10,

line(-5,0,-5.25,7.996092796),
line(-5.25,7.996092796,4.75,7.996092796),
line(5,0,4.75,7.996092796),
line(-5,0,5,0),
line(5,0,-5.25,7.996092796),
locate(-5.4,0,B), locate(5.1,0,C),locate(-5.72,8.37,A),
locate(4.85,8.4,D),locate(-.5,0,10),locate(-5.5,4.1,8),
locate(-.5,5,13)


 )}}}
<pre>
We use the Law of cosines to find &#8736;ABC

{{{matrix(1,9,


(matrix(2,1,THIRD,SIDE))^2,
"",
""="",
"",
(matrix(2,1,FIRST,SIDE))^2,
""+"",
(matrix(2,1,SECOND,SIDE))^2,
""-"",
2*(matrix(2,1,FIRST,SIDE))*(matrix(2,1,SECOND,SIDE))*(matrix(5,1,COSINE,OF,ANGLE,BETWEEN,THEM))) }}}

Chosse AC as the "THIRD SIDE"

AC² = AB² + BC² - 2·AB·BC·cos(&#8736;ABC)

13² = 8² + 10² - 2·8·10·cos(&#8736;ABC)
169 = 64 + 100 - 160·cos(&#8736;ABC)
169 = 164 - 160·cos(&#8736;ABC)
5 = -160·cos(&#8736;ABC)
{{{-5/160}}} = cos(&#8736;ABC)
-0.03125 = cos(&#8736;ABC)

So &#8736;ABC is an obtuse angle in the 
second quadrant. so we find the reference angle
by finding the inverse cosine of +0.03125 which is
88.20921534° and subtract from 180° and get 

&#8736;ABC = 91.79078466°

Since two adjacent angles of a parallelogram are 
supplementary, we can find &#8736;BAD by subtracting &#8736;ABC 
from 180°.

But we'll just get 88.20921534° again which is the same as
the reference angle for &#8736;ABC.

Therefore &#8736;BAD = 88.20921534°

To get that to the nearest minute we multiply the
decimal part 0.20921534 by 60 getting 12.55229204
which rounds to 13 minutes.

Answer: &#8736;BAD = 88°13'

Edwin</pre>