Ridiculously hard geometry problem

In △ADB,∠ADB=(180−18−30)∘=132∘

Applying sine law in △ADB,

ABsin132∘=ADsin30∘⟹AD=AB2sin48∘

as sin132∘=sin(180−132)∘=sin48∘ ∠ABC=∠BAC=180∘−96∘2=42∘

Applying sine law in △ABC, ACsin42∘=ABsin96∘⟹ACAB=sin42∘sin96∘=cos48∘2sin48∘cos48∘ (applying sin2A=2sinAcosA)

So, AC=AB2sin48∘⟹AC=AD So, ∠ACD=∠ADC=(180−24)∘2=78∘

Daniel Ettedgui its for you :))

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

I tried conventional methods to figure out this difficult problem but was unable to do so. When I stepped out of the box the solution jumped out at me. Enjoy. Triangle ABC has a mirror image, meaning at B the 42 degree angle is 18 + 24 degrees and at A it is 30 + 12 degrees. Let's name new triangle AB'C'. The corresponding point in the triangle is point D'. Point A is the same as we are superimposing new triangle over original triangle and rotating new triangle counterclockwise 60 degrees with point A as center of rotation. This will create a parallelogram D'B'CA. By symmetry, AD is parallel to C'B'. Also by symmetry, AC' is parallel to B'D. Triangle AB'D is an isosceles triangle with base angles of 42 degrees. Therefore AD = B'D' and triangles AC'D' and ADB' are congruent. Therefore AD' = C'B'. This means that AD = AC and therefore triangle ADC is isosceles with vertex angle of 24 degrees. So the base angles which include angle ACD are 78 degrees.

I just looked at the diagram it had originally (I can't see it anymore for some reason) but I looked at it and it just seemed like if I did 96-18 it'd make sense and oh wow it equals 78 look at that. To be honest I read the other solutions and I don't understand the math behind them at all, but it seemed that looking at the diagram you logically just had to do that simple subtraction and it would make sense. Sorry for bringing the average IQ of those who solved this down to 7 because I'm still only learning pre-calculus in high school. But I'd like to know (in simple terms please) how wrong I am because it was certainly just a lucky guess.