Find the angle inside a triangle in terms of two variables : $\alpha,\beta$

Question

In the adjoining figure(below)

CZ is perpendicular to XY and the ratio of AZ to ZB is $1 : 2$. The angle $ACX$ is $\alpha$ and the angle $BCY$ is $\beta$. Find an expression for the angle $AZC$ in terms of $\alpha$ and $\beta$.

By using the sine rule (in my opinion, which can be used here):

$\dfrac{AZ}{sin(90^{\circ}-\alpha)}=\dfrac{ZC}{sin\angle{ACB}}=\dfrac{AC}{sin\angle{AZC}}$.........(1)

$\dfrac{BZ}{sin(90^{\circ}-\beta)}=\dfrac{ZC}{sin\angle{ABC}}=\dfrac{BC}{sin({\pi-AZC})=sin\angle{AZC}}$..........(2)

Now $\dfrac{AZ}{ZB}=\dfrac{1}{2}$(Given)

$\dfrac{\dfrac{AZ}{ZC}}{\dfrac{ZB}{ZC}}=\dfrac{1}{2}$........(3)

Using $(1),(2) \space and \space (3)$, we get,

$\dfrac{AC}{BC}=\dfrac{sin\angle{ABC}}{sin\angle{CAB}}$= $\dfrac{1}{2}\cdot\dfrac{cos\beta}{cos\alpha}$

Again $\angle{CAB}+\angle{ABC}=\alpha+\beta$(by careful observation)

Again using cosine rule on $\Delta{ACZ}$, we get

$\sqrt{AC^2+ZC^2-2AC\cdot ZC \cdot sin \alpha}= AZ$

Putting value of AZ above in $(1)$, we get,

$\dfrac{\sqrt{AC^2+ZC^2-2AC\cdot ZC \cdot sin \alpha}}{cos\alpha}=\dfrac{AC}{sin\angle{AZC}}$

But here I am stuck as I think I am going somewhere else. Please give me any suggestion, idea or directly, the answer(if you'd actually do that I'll be obliged)

The probable solution is below:

Answer 1

Your solution looks good to me.

Answer 2

Here is the solution for the above question. Please see if there is some error.

In the 3rd line it should be angle $ZCA=90-\alpha$, wrongly written as $ZAC=90-\alpha$