Hey!

This is the solution:

We know that,

cosA= AB/ AC

cosX= XY/ XZ

Now, it is given that cosA= cosX

Therefore, AB/AC= XY/XZ

Let, these fractions be equal to k.

Therefore, AB= k.XY and AC= k.XZ ........(1)

Now, we need to find BC/ZY

So, BC/ZY= sqrt[AC^2- AB^2]/ sqrt[XZ^2- XY^2]

Putting the values from equation (1),

BC/ZY= k

Therefore, AB/XY= AC/XZ= BC/ZY

Therefore, the two triangles are similar.

And by property of similarity, angleA= angleX.

Hope this answer helps!