Question 667490
<pre>
Here is the triangle without the point D.

{{{drawing(300,400,-3,3,-1,7, locate(-.18,4.2,"20°"),
locate(-1.1,0,B), locate(1,0,C), locate(-.05,6,A),
triangle(-1,0,1,0,0,tan(80*pi/180))  )}}}

Before we put in the point D, let's chop the isosceles 
triangle into two right triangles with a median to the 
base, like this green line AE, and we will let BE = 1 unit
making BC = 2 units:

{{{drawing(300,400,-3,3,-1,7, locate(-.5,3,"10°"),
locate(-1.1,0,B), locate(1,0,C), locate(-.05,6,A), locate(0,0,E),
triangle(-1,0,1,0,0,tan(80*pi/180)),green(line(0,0,0,tan(80*pi/180))),
locate(-.6,0,1), locate(-.85,.35,"80°")  
  )}}}

{{{AB/BE}}} = sec(80°), {{{AB/1}}} = sec(80°), AB = sec(80°) = AC




{{{drawing(300,400,-3,3,-1,7, locate(-.18,4.2,"20°"),
locate(-1.1,0,B), locate(1,0,C), locate(-.05,6,A),
triangle(-1,0,1,0,0,tan(80*pi/180)), locate(-.55,4,D),
locate(-.4,4.8,2),  locate(0,0,2), locate(.6,3,"sec(80°)"),
line(-.3472963553,3.701666314,1,0)  )}}}

Now we have a case of Side-angle-side with triangle ADC.

So we use the law of cosines first to find the length of CD.

CD˛ = AD˛ + AC˛ - 2·AD·AC·cos(20°)

CD˛ = 2˛ + sec˛(80°) - 2·2·sec(80°)·cos(20°)

CD˛ = 4 + sec˛(80°) - 4·sec(80°)·cos(20°)

CD˛ = 15.51754097

CD = 3.939231012

Now we turn to the law of sines:

{{{CD/sin(A)}}} = {{{AC/sin(ADC)}}}

Cross multiply:

CD·sin(ADC) = AC·sin(A)

sin(ADC) = {{{(AC*sin(A))/(CD)}}}

sin(ADC) = {{{("sec(80°)"*sin("20°"))/(3.939231012)}}}

sin(ADC) = 0.5

< ADC = 150°

Edwin</pre>