Shift, Rotate, You're set

Label the diagram as follows:

Then $\angle DAE = \tan^{-1} \cfrac{DE}{AE} = \tan^{-1} \cfrac{3}{15} = \tan^{-1} \cfrac{1}{5} \approx 11.3099°$ , and by the Pythagorean Theorem on $\triangle ADE$ , $AD = \sqrt{AE^2 + DE^2} = \sqrt{15^2 + 3^2} = 3\sqrt{26}$ .

Now consider the original ellipse $\cfrac{x^2}{100} + \cfrac{y^2}{25} = 1$ :

The semi-major axis is $a = \sqrt{100} = 10$ and the semi-minor axis is $b = \sqrt{25} = 5$ , so the foci are at $(\pm \sqrt{a^2 - b^2}, 0) = (\pm \sqrt{10^2 - 5^2}, 0) = (\pm 5\sqrt{3}, 0)$ , which means $A'$ is at $A'(-5 \sqrt{3}, 0)$ .

If $D'$ is at $D'(D_x, D_y)$ , then by the distance formula $A'D' = \sqrt{(D_x - A_x)^2 + (D_y - A_y)^2} = \sqrt{(D_x + 5 \sqrt{3})^2 + D_y^2} = AD = 3\sqrt{26}$ , and since $D'$ is on the ellipse, $\cfrac{D_x^2}{100} + \cfrac{D_y^2}{25} = 1$ , which solves to $D'(D_x, D_y) = D'\bigg(\cfrac{6\sqrt{78}-20\sqrt{3}}{3}, -\sqrt{20\sqrt{26}-\cfrac{259}{3}}\bigg)$ .

That means $\angle D'A'B' = \tan^{-1} \cfrac{B'D'}{A'B'} = \tan^{-1} \cfrac{\sqrt{20\sqrt{26}-\frac{259}{3}}}{\frac{6\sqrt{78}-20\sqrt{3}}{3}} \approx 14.9863°$ .

The angle of rotation is then $\theta = \angle DAE + \angle DAB = \angle DAE + \angle D'A'B' \approx 11.3099° + 14.9863° \approx 26.2962°$ , so $\lfloor 100\theta \rfloor = \boxed{2629}$ .