<s>[INST] Finding a particular angle from a triangle 
The way I have seen a solution is very lengthy  to find the value  of x. Is there any easy way to get the value of x? I have tried   different formula and didn't get any shortcut  solution. 
 [/INST] 
Place point $E$ on $\overline{AD}$ such that $\angle CED = 60^{\circ}$.
Now, by the Law of Sines in $\triangle ABD$,
$$\frac{\sin 15^{\circ}}{1}=\frac{\sin 45^{\circ}}{\overline{AD}}$$
$$\therefore \overline{AD} = 1 + \sqrt{3}$$
Since $\triangle CDE$ is equilateral, $\overline{DE} = 2$.
$$\therefore \overline{AE} = \overline{AD} - \overline{DE} = (1 + \sqrt{3}) - 2 = \sqrt{3} - 1$$
Now notice:
$$\frac{\overline{AE}}{\overline{EC}} = \frac{\overline{BD}}{\overline{DA}}$$
Because:
$$\frac{\sqrt{3} - 1}{2} = \frac{1}{\sqrt{3} + 1}$$
Since that's true, and since $\angle BDA = \angle AEC = 120^{\circ}$, $\therefore \triangle ABD \sim \triangle CAE$.
The rest falls into place.
$\angle CAE = \angle ABD = 45^{\circ}$, so $\angle CAB = 15^{\circ} + 45^{\circ} = 60^{\circ}$, meaning $\angle ACB = \angle x = \color{red}{75^{\circ}}$.
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