I note that no angles have been given, so either they're not relevant or a piece of info is missing from the question. I'm going to assume that ABC is a right-angled triangle with the right angle at B. Then, applying Pythagoras' theorem, AC^2=AB^2+CB^2=144+25=169, so AC=sqrt(169)=13. In triangle DAC we can apply the cosine rule to find DC. DC^2=AD^2+AC^2-2*AD*ACcosDAC. Because AD and BC are parallel, DAC=ACB (angles), cosDAC=cosACB=5/13. So DC^2=33.8^2+169-2*33.8*13*5/13=1142.44+169-338=973.44 and DC=sqrt(973.44)=31.2. Apply the sine rule: 31.2/sinDAC=31.2/sinACB=31.2*13/12=33.8, which happens to be the length of AD, so sin90=1 because AD/sinDCA=33.8. Therefore DCA is a right angle and CDA=CAB. Therefore sinCAB=sinCDA=5/13 and CAB=CDA=22.62 degrees.