5. Find the smallest number $k$, such that $\frac{t_{a}+t_{b}}{a+b}<k$, where $a, b$ are two sides of a triangle, and $t_{a}, t_{b}$ are the lengths of the angle bisectors corresponding to these two sides.
5. When $\triangle A B C$ satisfies $a=b$, and the base angle $A \rightarrow 0$, then $c \rightarrow 2 b, \cos \frac{A}{2} \rightarrow 1$

Also, $t_{a}=\frac{b c \sin A}{(b+c) \sin \frac{A}{2}}=\frac{2 b c}{b+c} \cos \frac{A}{2} \quad$ so $\frac{t_{a}+t_{b}}{a+b}=\frac{t_{a}}{a} \rightarrow \frac{4}{3}$, hence $k \geqslant \frac{4}{3}$. On the other hand, $t_{a}2\left(\frac{2}{3}-\frac{b}{a+2 b}-\frac{a}{2 a+b}\right)=\frac{2(a-b)^{2}}{3(a+2 b)(2 a+b)} \geqslant 0$,
so $\frac{t_{a}+t_{b}}{a+b}<\frac{4}{3}$.
Therefore, the minimum value of $k$ is $\frac{4}{3}$.