10. Let $M$ be the set of all $\triangle ABC$ satisfying the conditions $BC=1, \angle A=2 \angle B$. If $\triangle ABC \in M$, and the angle bisector $CD$ of $\angle C$ divides the opposite side $AB$ into two segments in the ratio $AD:DB=t$, then $\triangle ABC$ is associated with the real number $t$. Determine the set of all such real numbers $t$.
10. Let $\triangle A B C \in M$. Draw the angle bisector $A E$ of $\angle A$, then!! According to the given, $\angle 1=\angle 2=\angle B$. Therefore, $\triangle A E C \sim \triangle B A C$. Combining with the properties of the angle bisector of a triangle,

we get $t=\frac{A D}{D B}=\frac{A C}{D C}=\frac{E C}{A C}=\frac{B E}{A B}$. Since $\triangle A E B$ is an isosceles triangle, we have $\cos B=\frac{\frac{1}{2} \cdot A B}{B E}=\frac{A B}{2 \overline{B E}}$, thus $t=\frac{1}{2 \cos B}$.

On the other hand, $\because 0^{\circ}<\angle B+\angle A<180^{\circ}$, $\angle A=2 \angle B, \therefore 0^{\circ}<3 \angle B<180^{\circ}$, i.e., $0^{\circ}<\angle B<60^{\circ}$. Therefore, the set of all such $t$ is $\left\{t \left\lvert\, t=\frac{1}{2 \cos \bar{B}}\right., \quad 0^{\circ}<\angle B<60^{\circ}\right\}$ $=\left(\frac{1}{2}, 1\right)$.