Using your result of $\measuredangle CQD = 90^{\circ} + \theta$ and the Law of sines with $\triangle CDQ$ gives
$$\begin{equation}\begin{aligned}\frac{\lvert DC\rvert}{\sin(90^{\circ} + \theta)} & = \frac{\lvert DQ\rvert}{\sin(2\theta)} \\\frac{\lvert DC\rvert}{\cos\theta} & = \frac{\lvert DQ\rvert}{2\sin\theta\cos\theta} \\\lvert DC\rvert & = \frac{\lvert DQ\rvert}{2\sin\theta}\end{aligned}\end{equation}$$
From $\triangle ADQ$, we get
$$\sin\theta = \frac{\lvert DQ\rvert}{\lvert AQ\rvert} \;\;\to\;\; \lvert DQ\rvert = 10\sin\theta$$
Substituting this into the earlier equation results in
$$\lvert DC\rvert = \frac{10\sin\theta}{2\sin\theta} = 5$$
However, this doesn't match your stated answer. In general, $\lvert DC\rvert$ is half that of $\lvert AQ\rvert$, as your diagram also roughly indicates.