2011-II-49

Since $DA$ is the tangent to the circle at $A$, then

$\begin{array}{rcl}\mathrm{\angle }DAC& =& \mathrm{\angle }ABC\\ & =& {28}^{\circ }\end{array}$

Since $BC$ is a diameter of the circle, then $\mathrm{\angle }CAB={90}^{\circ }$. Hence, we have

$\begin{array}{rcl}\mathrm{\angle }ACD& =& \mathrm{\angle }CAB+\mathrm{\angle }ABC\\ & =& {90}^{\circ }+{28}^{\circ }\\ & =& {118}^{\circ }\end{array}$

In $\mathrm{\Delta }ACD$,

$\begin{array}{rcl}\mathrm{\angle }ADB& =& {180}^{\circ }–\mathrm{\angle }ACD-\mathrm{\angle }DAC\\ & =& {180}^{\circ }–{118}^{\circ }–{28}^{\circ }\\ & =& {34}^{\circ }\end{array}$