Find the angle in the given figure

Question

Is question is a physics problem related to surface tension, but that isn’t relevant here.

Angle $CDB=\frac{\alpha}{2}$ and the angle $\theta$ are where they are as shown in the figure.

How is angle $ABC=\theta +\frac{\alpha}{2}$?

I couldn’t find any relation between those two, but that obtaining that result is imperative for solving this. How are they related?

Answer 1

Hint:

Extend the ray of $\theta$ from B to intersect AD at E. Triangles ABE and ABC are similar because they are ridght angle and have a common angle $\angle BAD$. Therefore $\angle BEA=\angle ABC$.But $\angle BEA=\theta +\frac {\alpha}2$ because it is external angle of triangle BDE. Therefore :

$\angle ABC=\angle BEA=\theta +\frac {\alpha}2$

Answer 2

Say the tangent to the liquid interface from $B$ meets $AD$ at $P$. Then considering the exterior angle of $\triangle BPD$, $$\angle BPC=\theta+\frac{\alpha}{2} \\ \implies \angle PBC=\frac{\pi}{2}-\left(\theta+\frac{\alpha}{2}\right) \\ \implies \angle ABC=\theta +\frac{\alpha}{2}$$