<s>[INST] Find the angle $\angle ACD=\theta$. Problem: Triangle $\triangle ABC$ has $AC=BC$ and $\angle ACB=96^{\text{o}}$. $D$ is a point in $\triangle ABC$ such that $\angle DAB=18^{\text{o}}$ and $\angle DBA=30^{\text{o}}$. Compute $\angle ACD.$
Drawing the problem first:

From the law of cosine we have that $b^2=2a^2-2a^2\cos{96}=2a^2(1-\cos{96})$ which leads to $b=|a|\sqrt{2(1-\cos{96})}.$
Since $\angle CAB=\angle ABC=(180-96)/2=42$, it follows that $\angle CAD=42-18=24$ and $\angle CBD = 42-30=12$. This information allows us to express $f$ in two ways by the law of cosine: \begin{array}{lcl}
f^2 & = & e^2+a^2+2ea\cos{12} \ \ \ \ \ \ \ \ \ \quad \ \quad (1)\\
f^2 & = & d^2+a^2+2da\cos{24} \ \ \ \ \ \ \ \ \ \quad \ \quad (2)\\
\end{array}
Subtracting (1)-(2) we get $e^2-d^2+2a(e\cos{12}-d\cos{24})=0$. 
Here I'm stuck. Any suggestion on how to proceed?
 [/INST] Using Law of Sines in triangle $\triangle{ABC}$ we can write:
$$
\frac{a}{\sin 42}=\frac{b}{\sin96} \Rightarrow b=\frac{a\sin 96}{\sin42}
$$
Similarly in triangle $\triangle{ABD}$ we have:
$$
\frac{b}{\sin 132}=\frac{d}{\sin30} \Rightarrow b=2d\sin 132=2d\cos42
$$
Equating $b$'s gives us:
$$a = \frac{2d\sin 42 \cos 42}{\sin 96}=\frac{d \sin84}{\cos6}=\frac{d\cos6}{\cos6} \Rightarrow a=d$$
So triangle $\triangle{ACD}$ is an isosceles triangle which yields $\measuredangle{ACD}=78^{\circ}$
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