Angle Chasing

Let $E$ be a point such that $BCED$ is a $||$ gm. Now, $BC = DE, \angle DCE = \angle BDC = 28^{\circ}$ and also, $\angle DEC = \angle DBC = 82^{\circ}$ . Therefore, $\angle DEC + \angle DAC = 180^{\circ} => ACED$ is a cyclic quadrilateral, also $BC = AD = DE => \angle ACD = \angle DEC = 28^{\circ}$

Let $BC=AD=1$ and $\angle ACD = \theta$ .

We note that $\angle BDC = 180^\circ - \angle DBC - \angle BCD = 180^\circ - 82^\circ - 70^\circ = 28^\circ$ . Using sine rule , we have:

$\begin{aligned} \frac {DC}{\sin \angle DBC} & = \frac {BC}{\sin \angle BDC} \\ \frac {DC}{\sin 82^\circ} & = \frac 1{\sin 28^\circ} \\ \implies DC & = \frac {\sin 82^\circ}{\sin 28^\circ} \end{aligned}$

Using sine rule again,

$\begin{aligned} \frac {\sin \angle ACD}{AD} & = \frac {\sin \angle DAC}{\color{#3D99F6}DC} & \small \color{#3D99F6} \text{Note that }DC = \frac {\sin 82^\circ}{\sin 28^\circ} \\ \frac {\sin \theta}1 & = \frac {\color{#D61F06}\sin 98^\circ \times \color{#3D99F6} \sin 28^\circ}{\color{#3D99F6}\sin 82^\circ} & \small \color{#3D99F6} \text{Note that } \sin (180^\circ -x) = \sin x \\ \sin \theta & = \frac {\color{#D61F06}\sin 82^\circ \times \sin 28^\circ}{\sin 82^\circ} \\ \sin \theta & = \sin 28^\circ \\ \implies \theta & = \boxed{28}^\circ \end{aligned}$