Originally posted by ThudanBlunder
Oops, sorry about that.
I meant AB = AC.
This was more painful than I thought it was going to be.
I have the angle as the inverse tangent of
( sin(20) * sin(30) * sin(60) ) / ( sin(20) * sin(80) - cos(20) * sin(30) * sin(60) ).
Here angles are in degrees.
I'm too tired at the moment to give a decimal approximation for this number. There's also a high chance of working error...basically I got all the angles I could from looking at triangles...but that still leaves 1 angle free. Then I used the sine rule lots to get that
sin(a+20) / sin(a) = ( sin(20) * sin(80) ) / ( sin(30) * sin(60) )
where a is the angle you're after.