Question

spaceblastxy1428

In $\bigtriangleup XYZ, \angle XYZ=70^\circ$ and $\angle XZY=50^\circ.$ Points $A$ and $B$ lie on sides $\overline{XY}$ and $\overline{XZ}$ such that $\angle AZY=40^\circ$ and $\angle BYZ=50^\circ.$ Find $\angle ABY.$

How do you find the answer? $($There is only one answer because there is only one triangle with angles $70^\circ \text{ and } 50^\circ,$ there is only one point on $\overline{XY}$ so that $\angle AZY=40^\circ,$ and there is only one point on $\overline{XZ}$ so that $\angle BYZ=50^\circ.)$

debbie

This post is deleted!

debbie

@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.