9. Let $x=0.82^{0.5}, y=\sin 1, z=\log _{3} \sqrt{7}$. Then the size relationship of $x, y, z$ is ( ).
(A) $x<y<z$
(B) $y<z<x$
(C) $z<x<y$
(D) $z<y<x$
9. B.

From $x=0.82^{0.5}>0.81^{0.5}=0.9, y=\sin 17^{5}$ we get $9>5 \log _{3} 7$, which means $0.9>\log _{3} \sqrt{7}$, so $x>z$.
Also, from $3^{7}<7^{4}$, we can get $3^{\frac{7}{4}}<7$.
And $3^{\sqrt{3}}<7, \sqrt{3}<\log _{3} 7, \frac{\sqrt{3}}{2}<\log _{3} \sqrt{7}$, which means $y<z$.