@spaceblastxy1428 Thanks so much for updating this question with more info!
I tried it, and I'm sorry that I don't have an elegant way to do it, just a grungy way
Here is the grungy way:
Angle-chasing (which we learned about in these lessons: M2W2D5-Y-2, M2W4D13-Ch-2) tells us
that \( \angle BYZ = 50^{\circ} \) and \(\angle YAB = 70^{\circ},\) so since \( \angle BZY = 50^{\circ}\) and \(\angle AYZ = 70^{\circ}\) (which is given), we know that triangle \(\bigtriangleup BYZ\) and \(\bigtriangleup AYZ\) are both isosceles.
I'll let \( BZ = BY = 1,\) since all lengths will be relative and only the angles matter, and normalize all other lengths around this length. Then triangle \(\bigtriangleup BYZ\) has lengths of \(1, 1,\) and a third leg, the length of which can be found using the Law of Sines:
$$ \frac{\sin 50^{\circ}}{1} = \frac{\sin 80^{\circ}}{YZ}$$
$$ YZ = \frac{\sin 80^{\circ}}{\sin 50^{\circ}} $$
From angle-chasing, we know that \( \overline{AZ} \) and \(\overline{YB}\) intersect at right angles. Let's call this intersection point \(D.\) Then we get four right triangles: \( \bigtriangleup DYZ, \bigtriangleup DAY, \bigtriangleup DBZ, \text{ and } \bigtriangleup DAB.\)
Then \(BD + DY = 1\) and \(AD + DZ = \frac{\sin 80^{\circ}}{\sin 50^{\circ}}.\)
From the Pythagorean Theorem on \(\bigtriangleup DBZ,\) we know \(BD^2 + DZ^2 = 1,\) so \(DZ^2 = 1 - DB^2.\)
From the Pythagorean Theorem on \(\bigtriangleup DYZ,\) we know that \(DY^2 + DZ^2 = \left( \frac{\sin 80^{\circ}}{\sin 50^{\circ}} \right)^2 .\)
Substitute for \(DZ^2\) into the first equation to solve for \(DB,\) which equals \( 1 - \frac{1}{2} \left( \frac{\sin 80^{\circ}}{\sin 50^{\circ}} \right)^2.\)
Then solve for \(DZ\) using the Pythagorean Theorem on \(\bigtriangleup DBZ,\) which says \( DZ^2 = 1 - DB^2.\)
Once you have found \(DZ,\) then use it to find \(AD,\) since \(AD + DZ = \frac{\sin 80^{\circ}}{\sin 50^{\circ}}.\)
Once you have found \(AD,\) then using the value for \(DB\) from earlier, we can solve for \(\angle ABY:\)
$$ \tan ABY = \frac{AD}{DB} $$
$$ \angle ABY = \arctan \frac{AD}{DB} $$
Sorry I didn't do the calculations... they were just so grungy! But the logic of the steps is all here.