Angles in sequence

Let the angles be $\alpha,90-\alpha,90^{\circ}$ .

If they are in arithmetic sequence , then $(90-\alpha)-\alpha=90-(90-\alpha)~~ \color{#3D99F6}\text{(because common difference is same)}$ $\implies 90-2\alpha=\alpha \implies \alpha=30^{\circ}=x$ .

If they are in geometric sequence , then $\dfrac{90-\alpha}{\alpha}=\dfrac{90}{90-\alpha} ~~ \color{#3D99F6}\text{(because the common ratio is same)}$ $\implies 90\alpha=(90-\alpha)^2$

$\implies 90\alpha=8100-180\alpha+\alpha^2 \implies \alpha^2-270\alpha+8100=0$ .

Which on solving gives $\alpha = 135\pm 45\sqrt{5}$ , since the angle is acute we only consider $\alpha=135-45\sqrt{5} \approx 34.4^{\circ}=y$

$\implies y-x \approx 34.4-30 \approx 4.3$ and nearest integer to it is $\boxed{4}$