In a triangle $ABC$, $AB = a-b$ and $BC = 2\sqrt{ab}$, then find $\angle B$? Is this question solvable?

In $\Delta ABC$, $AB = a-b$ and $BC = 2\sqrt{ab}$, then $\angle B$ is
(a) $\: 60^{\circ}$
(b) $\: 30^{\circ}$
(c) $\: 90^{\circ}$
(d) $\: 45^{\circ}$


Using @JanEerland ‘s suggestion, $BC = k \sin A$, $AC = k \sin B$, and $AB = k \sin C$; for some $k \ne 0$.
Substituting in the cosine law (wrt B), we get
$$\sin^2 B = \sin^2 A + \sin^2 C – 2 \sin A \sin C \cos B$$
(There might be others) but one solution of it is $B = 90^0$ and $A$ is then complement to $C$.
Note that, up to this point, the given $AB = a - b$ and $BC = ...$ have not been used. They probably are used to find $AC = … = a + b$ by Pythagoras theorem, and for the checking of triangle inequality (which maybe or maybe not necessary). This is not difficult to do with the help of $a – b \gt 0$.
However, the notation used in the question is very misleading. This is because according to our common naming convention, for $\triangle ABC$, $BC = a$, $AC = b$ etc. Such confusion can be found in various comments. 
