find the alpha angle

Join $BD$ . Let $CD = CB = BA = s$ .

In $\triangle BCD$ ,

$\angle CDB + \angle CBD = 180\degree - 96\degree = 84\degree$

$\Rightarrow 2\angle CDB = 84\degree$

$\Rightarrow \angle CDB = \angle CBD = 42\degree$

$\Rightarrow ADB = 54\degree - 42\degree = 12\degree$

Now

$BD = s\text{ cos}(\angle CDB) + s\text{ cos}(\angle CBD)$

$\Rightarrow BD = 2s\text{ cos}(42\degree)$

In $\triangle ABD$ , using sine rule

$\Large\frac{\text{sin}(\angle DAB)}{BD}$ $= \Large\frac{\text{ sin}(\angle ADB)}{AB}$

$\Rightarrow \Large\frac{\text{sin}(\alpha)}{2s\text{ cos}(42\degree)}$ $= \Large\frac{\text{ sin}(12\degree)}{s}$

$\Rightarrow \text{sin}(\alpha) = 2\text{ cos}(42\degree)\text{ sin}(12\degree)$

$\Rightarrow \text{sin}(\alpha) = \text{ sin}(54\degree) - \text{ sin}(30\degree)$

$\Rightarrow \text{sin}(\alpha) = 3\text{ sin}(18\degree) - 4\text{ sin}^3(18\degree) - \text{ sin}(30\degree)$

Putting $\text{ sin}(18\degree) = \Large\frac{\sqrt{5} - 1}{4}$ and $\text{ sin}(30\degree) = \Large\frac{1}{2}$ , we get

$\text{sin}(\alpha) = \Large\frac{\sqrt{5} - 1}{4}$ $= \text{sin}(18\degree)$

$\Rightarrow \alpha = 18\degree$