In a triangle ABC , with usual notations, if $a=5, \mathrm{~b}=7, \sin \mathrm{~A}=\frac{3}{4}$, then total number of triangles possible are

Using the Law of Sines with sides $a = 5$, $b = 7$, and $\sin A = \frac{3}{4}$, we establish the relation:

$\frac{a}{\sin A} = \frac{b}{\sin B}$

This gives $\frac{5}{\frac{3}{4}} = \frac{7}{\sin B}$

Solving for $\sin B$:

$\sin B = \frac{7 \times \frac{3}{4}}{5} = \frac{21}{20}$

Since $\frac{21}{20} > 1$ exceeds the valid range of sine values $[-1, 1]$, no such angle $B$ exists.

Thus, the number of possible triangles is $\boxed{0}$.