90 Given that the three sides $a, b, c$ of $\triangle A B C$ form a geometric sequence, and the opposite angles of $a, b, c$ are $\angle A, \angle B, \angle C$ respectively, then the range of $\sin B+\cos B$ is
A. $\left[\frac{1}{2}, 1+\frac{\sqrt{3}}{2}\right]$
B. $(1, \sqrt{2}]$
C. $\left(1,1+\frac{\sqrt{3}}{2}\right]$
D. $\left[\frac{1}{2}, \sqrt{2}\right]$
90 B. Let the three sides be $1, q, q^{2}(q \leqslant 1)$. The side opposite B is $q$. It is easy to see that when $q \rightarrow \frac{\sqrt{5}-1}{2}$, $B \rightarrow 0^{\circ} ; q=1$ when, $B=60^{\circ}$.
It is well known that
$$
\cos B+\sin B>\cos ^{2} B+\sin ^{2} B=1
$$

and as $B \rightarrow 0^{\circ}$, $\cos B+\sin B \rightarrow 1$.
Also,
$$
\sin B+\cos B \leqslant \sqrt{2} \sin \left(45^{\circ}+45^{\circ}\right)=\sqrt{2},
$$