6. If the three sides of $\triangle A B C$ are $a=\sin \frac{3}{4}$, $b=\cos \frac{3}{4}$, $c=1$, then the size order of $\angle A, \angle B, \angle C$ is ( ).
(A) $\angle A<\angle B<\angle C$
(B) $\angle B<\angle A<\angle C$
(C) $\angle C<\angle B<\angle A$
(D) $\angle C<\angle A<\angle B$
6. A.

Since $\frac{3}{4}\cos \frac{3}{4} > \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$.

Then $a < b < c$. Hence $\angle A < \angle B < \angle C$.