The (other) Third Law

By the Law of Tangents ,

$\dfrac{\tan \frac{A - B}{2}}{\tan \frac{A + B}{2}} = \dfrac{a - b}{a + b}.$

Plugging in the values from the problem, we get

$\begin{aligned} \dfrac{\tan {\frac{45}{2}}^{\circ}}{\tan \frac{A + B}{2}} &= \dfrac{20 - 18}{20 + 18} \\ \dfrac{\sqrt{2} - 1}{\cot \frac{C}{2}} &= \dfrac{1}{19} \\ \tan \dfrac{C}{2} &= \dfrac{1}{19(\sqrt{2} - 1)} \\ \tan \dfrac{C}{2} &\approx 0.1270\dots \end{aligned}$

Thus, $\left \lfloor 1000 \tan \dfrac{C}{2} \right \rfloor = \boxed{127}.$

From $A-B=45$ , we get $A=45+B$ $\implies$ $\boxed{1}$

By sine law, we have

$\dfrac{\sin A}{20}=\dfrac{\sin B}{18}$ $\implies$ $\sin A=\dfrac{20}{18} \sin B$

Substitute $\boxed{1}$ in the above, we have

$\sin (45 + B) = \dfrac{10}{9} \sin B$

Use the identity: $\sin (\theta + \phi)=\sin \theta \cos \phi + \cos \theta \sin \phi$ , we have

$\sin 45 \cos B + \cos 45 \sin B = \dfrac{10}{9} \sin B$

$\sin 45 \cos B = \dfrac{10}{9} \sin B - \cos 45 \sin B=\sin B \left(\dfrac{10}{9}-\cos 45\right)$

Use the identity: $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$

$\tan B=\dfrac{\sin 45}{\dfrac{10}{9}-\cos 45}$

$B=\tan ^{-1} \left[\dfrac{\sin 45}{\dfrac{10}{9}-\cos 45}\right] \approx 60.25858149$

It follows that, $A=45+B \approx 105.2585815$

Since the sum of the interior angles of a triangle is $180$ , we get, $C=180-A-B \approx 14.48283701$

Finally, the desired answer is $\left\lfloor 1000 \tan \dfrac{C}{2} \right \rfloor = 1000 \times \tan \dfrac{14.48283701}{2}=\large{\boxed{127}}$