*Originally posted by ThudanBlunder*

**Oops, sorry about that.
**

I meant AB = AC.

ðŸ˜³

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.