Question 1162398
<pre>
The sum of the interior angles of an n-sided polygon is

SUM = (n-2)∙180°

So for a pentagon, the sum is

SUM = (5-2)∙180° = 3∙180° = 540°

Since all interior angles of a regular pentagon are equal, we
divide that by 5, and get 540°÷5 = 108°

So each of the interior angles of the pentagon measures 108°.

Since ∠B = 108° and ΔABC is isosceles, its base angles are 36° each, and the
same for ΔAED.  

Since all three angles at A add to 108°, ∠DAC also = 36°.

Since ΔADC is isosceles, its base angles are equal and 72° each.

Now you have all the angles of ΔADC.

{{{drawing(400,400,-1.6,1.6,-1.6,1.6,
line(0.95105652,0.30901699,0,1),
line(0,1,-0.95105652,0.30901699),
line(-0.95105652,0.30901699,-0.58778525,-0.80901699),
line(-0.58778525,-0.80901699,0.58778525,-0.80901699),
line(0.58778525,-0.80901699,0.95105652,0.30901699),
locate(-.02,1.13,A),locate(-1.05,.37,B),locate(-.65,-.8,C),

locate(-.9,.41,108^o), locate(-.25,.9,36^o), locate(.1,.85,36^o),
locate(-.67,-.35,36^o), locate(-.1,.79,36^o),

locate(.63,.41,108^o),  locate(.3,-.6,72^o),locate(-.55,-.6,72^o),

locate(.5,-.35,36^o),


line(0.58778525,-0.80901699,0,1),
line(-0.58778525,-0.80901699,0,1),

locate(1,.37,E), locate(.61,-.8,D)) }}}


Edwin</pre>