79 Let $a=\cos \frac{3}{2}, b=\sin \frac{1}{10}$, then the relationship between $a, b$ is
A. $a>b$
B. $a<b$
C. $a=b$
D. Cannot be determined
79 B. Since $a=\cos \frac{3}{2}=\sin \left(\frac{\pi}{2}-\frac{3}{2}\right)$,
and because
$$
\frac{\pi}{2}-\frac{3}{2}<\frac{1}{10},
$$

and $y=\sin x$ is an increasing function, so
$$
a=\sin \left(\frac{\pi}{2}-\frac{3}{2}\right)<\sin \frac{1}{10}=b .
$$