Given the right triangle XYZ
[tex]\begin{gathered} XZ^2=XY^2+YZ^2 \\ YZ^2=XZ^2-XY^2 \\ YZ^2=9^2-3^2=81-9=72 \\ YZ=\sqrt[]{72}=\sqrt[]{36\cdot2}=6\sqrt[]{2} \end{gathered}[/tex]We will find the angles using the sine rule
[tex]\begin{gathered} \angle Y=90,XZ=9 \\ \\ \frac{XZ}{\sin Y}=\frac{YZ}{\sin X}=\frac{XY}{\sin Z} \\ \\ \frac{9}{\sin90}=\frac{6\sqrt[]{2}}{\sin X}=\frac{3}{\sin Z} \end{gathered}[/tex]So, the angle X will be:
[tex]\begin{gathered} \frac{9}{\sin90}=\frac{6\sqrt[]{2}}{\sin X} \\ \\ \sin X=\frac{6\sqrt[]{2}}{9}\cdot\sin 90=0.9428 \\ \\ X=\sin ^{-1}0.9428\approx70.5^o^{} \end{gathered}[/tex]The angle Z will be:
[tex]\begin{gathered} \frac{9}{\sin90}=\frac{3}{\sin Z} \\ \\ \sin Z=\frac{3}{9}\sin 90=0.3333 \\ \\ Z=\sin ^{-1}0.3333\approx19.5^o \end{gathered}[/tex]So, the answer is:
[tex]\begin{gathered} YZ=6\sqrt[]{2}\approx8.5 \\ \\ \angle X=70.5^o \\ \\ \angle Z=19.5^o \end{gathered}[/tex]