### $\alpha = ?$ in the Triangle

Using Sine Rule in the left triangle,

$\frac {sin \alpha}{a} = \frac {sin \beta}{b}$

Using Sine Rule in the right triangle,

$\frac {sin 20^\circ}{a} = \frac {sin 140^\circ}{b}$

Dividing the above equations we have,

$\frac {sin \alpha}{sin 20^\circ}=\frac {sin \beta}{sin 140^\circ}$

From the left triangle, $\alpha + \beta + 40^\circ = 180^\circ$

$\Rightarrow \beta = 180^\circ - (\alpha +40^\circ)$

Thus, $\frac {sin \alpha}{sin 20^\circ}=\frac {sin[180^\circ-(\alpha+40^\circ)]}{sin (180-40)^\circ}$

Or, $\frac {sin \alpha}{sin 20^\circ}=\frac {sin (\alpha+40^\circ)}{sin 40^\circ}$

$\Rightarrow sin \alpha .2sin 20^\circ cos 20^\circ = sin 20^\circ (sin \alpha cos40 ^\circ +cos\alpha sin40^\circ )$

$\Rightarrow 2sin \alpha cos 20^\circ = sin \alpha cos40 ^\circ +cos\alpha sin40^\circ$

$\Rightarrow 2 cos 20^\circ = cos40 ^\circ +cot\alpha sin40^\circ$

$\Rightarrow cot \alpha = \frac{2cos 20 ^\circ - cos 40 ^\circ}{sin 40 ^\circ}$

Or, $cot \alpha = \frac {2cos (30^\circ-10 ^\circ) - cos (30^\circ+10 ^\circ)}{sin (30^\circ+10 ^\circ)}$

$\Rightarrow cot\alpha = \frac{{2\left( {\frac{{\sqrt 3 }}{2}\cos 10^\circ + \frac{1}{2}\sin 10^\circ } \right) - \left( {\frac{{\sqrt 3 }}{2}\cos 10^\circ - \frac{1}{2}\sin 10^\circ } \right)}}{{\frac{1}{2}\cos 10^\circ + \frac{{\sqrt 3 }}{2}\sin 10^\circ }}$

$\Rightarrow cot\alpha = \frac{{\frac{{\sqrt 3 }}{2}\cos 10^\circ + \frac{3}{2}\sin 10^\circ }}{{\frac{1}{2}\cos 10^\circ + \frac{{\sqrt 3 }}{2}\sin 10^\circ }} = \sqrt 3$

$\therefore \alpha = 30^\circ$