I've drawn the below diagram.
The circle has the radius $a$.
$$ \omega_{} \approx \frac{ \pi a ^{2} \cdot \cos^{}\left(\theta_{} \right) }{ r ^{2} } \tag{1} $$ $$a \ll r$$ I viewed diagrams of solid angle and noticed that a aquared euclid distance comes to a denominator and a numerator takes a sub-area of a sphere(not nessecary to be a sub-area of a sphere?) .
So at least I can get that the approximate equation $(1)$ takes $r ^{2}$ as the denominator but how $\pi a ^{2} \cdot \cos^{ }\left(\theta_{} \right) $ comes from? Moreover I've been unable to connect relation(s) between the approximate equation $(1)$ and the approximation? of $a \ll r$.
$\pi a ^{2}$ is one of the factors of the numerator and I can intuitevely get this factor can be one of the components of the some area.
Let $f:\mathbb R^3 \setminus \{0\} \to S^2$ be the projection onto the unit sphere:
$$f(v) = \frac v{\lVert v \rVert}.$$ Then $$(Df)_v(\varepsilon) = \frac{\pi_v(\varepsilon)}{\lVert v\rVert}$$ for some small offset $\epsilon$, where $\pi_v$ is the orthogonal projection onto $v^{\perp}$. This means that $f$ changes the area of a surface through $v$ having a surface normal $n$ at $v$ locally by a factor of $$ \frac {\langle n, \frac v{\lVert v\rVert} \rangle}{\lVert v \rVert^2}.$$ Note that the numerator is the cosine of the angle between $n$ and $v$. For your example this factor is $$\frac{\cos(\alpha)}{r^2}$$ and the area of the disc is $\pi a^2$. Together, this leads to your estimation of the solid angle.