Background. The matching number $m(G)$ of a graph $G$, is the size of a maximum independent edge set, and the vertex cover number $\tau (G)$ is the size of a minimum vertex cover of $G$. It is straightforward to see that $m(G)\leq \tau (G)$ for every graph $G$, and moreover the difference $\tau(G)- m(G)$ can be arbitrary high. For example, for the complete graph $K_n$, we have $m(K_n)= \lceil\frac{n}{2}\rceil$ while $\tau (K_n)=n-1$.
Question. Can this difference $\tau(G)- m(G)$ be arbitrary high when $G$ is a connected triangle-free graph? (For example, the odd cycles show the gap can be one.)
More generally, is it known the best upper bound for $\tau(G)$ in terms of $m(G)$ for this special class of graphs (triangle-free graphs)?
For the first question, one possible answer is as follows. Let $C_{2m+1}$ is a odd cycle of length $2m+1$ where $m\geq 2$. Then, construct $G$ as follows: To any vertex of the cycle attach a cycle $C_5$ of length $5$ in a way that all the new cycles being vertex disjoint pairwise. Then $\tau(G)\geq (2m+1)\times (3)$ (as any vertex cover contains at lease 3 vertex of each C_5) while $m(G)\leq m+ (2m+1)(2)$ (any matching set contains at most $m$ edges from the original cycles and 2 from each $C_5$). Thus, $m(G)-\tau(G)\geq m+1$.