Law of Sines will give a unique solution iff a > b? Given a triangle ABC, with known sides a=BC and b=AC, and known angle A,
we wish to find angle B.
This is a typical application of the Sine Rule (Law of Sines).
In some circumstances, the sine rule gives an ambiguous result: with two possible solutions for angle B.
I am trying to find the simplest way of identifying whether or not the sine rule would give a unique solution.
Is it true to say that the sine rule will give a unique solution to this problem iff a > b?
 A: The question isn't really (or shouldn't be) about the sine rule, but about when two sides and an angle not formed by the two sides determine a unique triangle. You found almost the right criterion; in fact if $a=b$ the triangle is also uniquely determined (unless you allow degenerate triangles). The sine rule, by contrast, always allows two different angles at $B$, since the sine is symmetric with respect to reflection at $\pi/2$. If $a\ge b$, you can exclude the greater of the two because the sum with the angle at $A$ would exceed $\pi$, whereas for $a\lt b$ both of these angles correspond to triangles.
A: This is also known as the ambiguous case
(or SSA), and occurs whenever $a>b\,\sin A$,
i.e., whenever $B\ne90^\circ$ and $b$ is not the hypotenuse
(and $a$ the leg) of a right triangle.

Angle $B$ can be acute,
$B_1=\arcsin\left(\frac{b}{a}\sin A\right)$,
or obtuse, $B_2=\pi-B_1=180^\circ-B_1$,
corresponding to which, side $c$
will be bigger ($c_1$) or smaller ($c_2$)
than $b\,\cos A$, the common length from the right triangle.
A: That is correct. (Or almost correct: there is no solution at all if $a < b \sin A$.)
