Answer
a. $ \angle {BCD} \approx 91.16^{\circ}$ or, $92^{\circ}$
b. $\angle {DCA}\approx 14.42^{\circ}$ or, $14.4^{\circ}$
Work Step by Step
a. Need to find the $\angle {BCD}$.
Use law of sines formula.
$\dfrac{\sin 30^{\circ}}{20}=\dfrac{\sin \angle B}{28}$
Then, we have $\sin \angle B=\dfrac{28 \sin 30^{\circ} }{20^{\circ}}=0.7$
This implies that $\angle {ABC}=44.4^{\circ}$
It can be seen that the triangle BCD is an equilateral triangle, so $BC=CD=20$. This means that $\angle {CDB}=\angle {CBD}=44.4^{\circ}$
Now, $\angle {CDB}+\angle {CBD}+\angle {BCD}=180^{\circ}$
$\implies \angle {BCD} \approx 91.16^{\circ}$
b. Need to find the $\angle {DCA}$.
$\angle {A}+\angle {B}+\angle {ACB}=180^{\circ}$
$\implies \angle {ACB} \approx 105.58^{\circ}$
Thus, $\angle {DCA}=\angle {ACB}- \angle {BCD}$
$\implies \angle {DCA}= 105.58^{\circ}-91.16^{\circ}\approx 14.42^{\circ}$
