Suspicious Quadrilateral

Construct equilateral $\triangle ACF$ . $\overleftrightarrow{CD}$ intersects $\overline{AF}$ at $E$ .

$m\angle CAF = 60°$ → $m\angle DAF = 48° = m\angle DAB$

$m\angle CAB = 36°$ , $m\angle ABC = 72°$ → $m\angle ACB = 72° = m\angle ABC$ → $AB = AC = AF$ ( $AC = AF$ because of equilateral $\triangle ACF$ )

$\triangle ABD \cong \triangle AFD\ (S.A.S.)$

$m\angle BAD = m\angle ABD$ → $AD = BD$ → $AD = FD$

$AD = FD$ and $AC = FC$ → $\overline{CE}$ is the bisector of $\angle ACF$ → $m\angle ACD = \fbox{30°}$