How do I find DB in this triangle?

You can use law of sines. Note that ADC=105 and hence ACD=45. Using law of sines in the triangle ADC: 

\[\frac{CD}{\sin 30}=\frac{AD}{\sin 45} \Rightarrow      CD=AD \frac{\sin 30}{\sin 45}=2\frac{\sin 30}{\sin 45}\]

Using law of sines in the triangle BDC: 

\[\frac{CD}{\sin 45}=\frac{BD}{\sin 60} \Rightarrow        CD=BD \frac{\sin 45}{\sin 60}.\]
Hence
\[CD=2\frac{\sin 30}{\sin 45}=BD \frac{\sin 45}{\sin 60}\]
\[\Rightarrow BD=2\frac{\sin 30 \sin 60}{(\sin45)^2}=2\frac{\frac{1}{2}\frac{\sqrt{3}}{2}}{(\frac{\sqrt{2}}{2})^2}=\sqrt{3}.\]