Right triangle A with hypotenuse 1 and sides x and y is graphed at the origin. Right triangle B is graphed so that it shares a hypotenuse with triangle A and it's sides w and z are parallel to the functions y = x and y = -x. Find w and z with respect to x and y.
A link to a visual of the problem
In the configuration of the visual, I've been able to figure out that z = cos(135 - arctan(y/x)) * sqrt(x^2 + y^2) and w = cos(45 - arctan(x/y) * sqrt(x^2 + y^2). HOWEVER, in another configuration where say, x is larger than y, or ANY of the coordinates were negative, this would not work.
I need a formula to find w and z that works 100% of the time.
Any and all help solving this problem would be appreciated.
We can start by labelling some angles in your picture.
https://i.stack.imgur.com/s6y6j.pngAssume that every angle that is on there is exactly like the picture you sent. This means that $\angle BAC = \sin^{-1} \left( y \right)$. Since the side length $w$ is on $y = -x$, it means that $\angle DAC = 135^{\circ}$. This means that $\angle DAB = 135^\circ - \sin^{-1} \left( y \right)$ Now we can use law of sines: $$\frac{1}{\sin 90^\circ} = \frac{z}{\sin \left(135^\circ - \sin^{-1} \left( y \right) \right)}$$ This means that $z = \sin \left( 135^\circ - \sin^{-1} \left( y \right) \right)$
Since we have two of the three angles in $\triangle DAB$ we know that $\angle DBA = 180 - \left(90 + \left(135 - \sin^{-1} \left( y \right) \right) \right) = 90 - \left( 135 - \sin^{-1} \left( y \right) \right)$. We repeat law of sines: $$\frac{1}{\sin 90^\circ} = \frac{w}{\sin \left(\angle DBA \right)} = \frac{w}{\sin \left( 90 - \left( 135 - \sin^{-1} \left( y \right) \right) \right)}$$ This means that $w = \sin \left(-45 - \sin^{-1}\left(y \right) \right)$.