Measure of an Interior Angle

ΔABC is a isosceles triangle, solve for angle CAB

CAB = (180-96)/2 = 42 , ADB=180-18-30=132 and CAD=42-18=24

Assume AC and CB equal to 1 and solve for AB by sine rule. Use AB to solve for DA by sine rule. Then we found that DA=AC, ΔACD is isosceles triangle, therefore ACD=(180-24)/2 = 78

I have a pure geometry solution.

It is apparent that CAD=24 and CBD =12. Take bisector of CAD and extend BD to let it meet at P.

Now CAP =12 and CBP = 12 . So AP=PB( PAB = PBA = 30 ^).

So P is on perpendicular bisector of AB and C is on perpendicular bisector of AB. So join CP. => ACP =96/2=48.

In triangle PAB , PA=PB => PAB= PBA=30 => APB=180-60=120. In triangle PAD PAD=12, APD=120 =>PDA=48. So triangle ACP and triangle ADP are congruent
(PCA=PDA=48,PAD=PAC=12,APC=APD=120 and SIDE AP is common)

APC <--->APD is congruent so AC=AD and So angles ACD=ADC= (180-24)/2=78.

Applying the cosine rule to $\angle ACB$ with $AC=BC$ , we have $\begin{aligned} AB &= AC \sqrt{2 (1- \cos 96^\circ)}\\ &= 2 (\sin 48^\circ) AC && \cos 96^\circ = 1 - 2 \sin^2 48 \\ \end{aligned}$

We also have that $\sin \angle ADB = \sin (180^\circ - 18^\circ - 30^\circ) = \sin 132^\circ = \sin 48^\circ$ . Thus applying the sine rule to triangle $ADB$ , and substituting for $AB$ , we have $DA = \frac {AB}{\sin 48^\circ} \times \sin 30^\circ = AC$ . Thus $DAC$ is an isosceles triangle.

Since triangle $ACB$ is isosceles, we have $\angle CAB = \frac {1}{2} (180^\circ - 96 ^\circ) = 42^\circ$ . Thus $\angle CAD = 42^\circ - 18^\circ =24^\circ$ . Now since $ACD$ is isosceles, we have $\angle ACD = \frac {1}{2} (180^\circ - 24^\circ) = 78^\circ$ .

Constructions:- Let $AD \cap CB = E.$ Draw the angle bisector of $\angle BAD$ to meet $BD$ produced at $I$ . Join $CI$ and $EI$ as well. It's easy to conclude via simple angle chasings that $\angle EAI = \angle EBI = 12^{\circ}$ and $\angle AEC = 60^{\circ}$ . Now in quadrilateral $AIEB$ , $\angle EAI = \angle EBI = 12^{\circ}$ , therefore quad. $AIEB$ is cyclic. $\angle DIE = \angle DAB = 18^{\circ} ---- eq^{n} 1$ Also, $\angle IEA = \angle IBA = 30^{\circ}.$ But since, $\angle AEC = 60^{\circ}$ and $\angle IEA = 30^{\circ}$ therefore, $EI$ is the bisector of $\angle AEC$ $=> I$ is the incenter. Therefore, $AI$ bisects $\angle A => \angle IAE = 48^{\circ}.$ But, $\angle IDE = \angle ADB = 132^{\circ}$ . Therefore, $\angle IAE + \angle EDI = 180^{\circ} => CIDE$ is cyclic. $=> \angle DCE = \angle DIE = 18^{\circ}$ [Using $eq^{n} 1$ ] $=> \angle ACD = 78^{\circ}.$

Just $K.I.P.K.I.G$

180 / 2 = 42. 42 - 18 = 24. 180- 24= 156 156/2 = 78. 78= angle ACD