There is a formula of the angle (say, $\phi$) between a plane $\pi$ and a line $l$:
$$\phi=\sin^{-1}\frac{\vec{b}.\vec{n}}{|\vec{b}||\vec{n}|}\tag{1}$$
Where $l$ is parellel to $\vec{b}$ and $\vec{n}$ is the normal vector to $\pi$ from the origin
I tried to convert it into a 2D problem and derive it in the following manner:
Let the required angle be $\theta$
The equation of line is $\vec{r}=\vec{a}+\lambda\vec{b}$ and that of the plane is $\vec{r}.\hat{n}=d$
Note: Here $\hat{n}$ is the normal unit vector to plane from the origin
In the figure, $AB$ lies on the plane $\pi$ and $\vec{AE}$ lies on $l$
$\vec{CB}$ is perpendicular distance to the plane from the origin, and is equal to $d\hat{n}$, also let $\vec{BE}=d\hat{n}$ where $d$ is some real number
Let's drop a perpendicular from a point $D$ to a point $A$ on the plane such that $AB$ and $DC$ are parallel
Clearly, the angle $BDC$ will also be $\theta$. Let $BD=x$
Then $BC=x\sin\theta$
$$x=\frac{d}{\sin\theta}$$
the angle $ABD$ will also be $\theta$
Therefore, $AB=BD\sin\theta$
$AB=d$
Now, we have to find the angle between $l$ and $AB$
Considering $l$ is parallel to $\vec{b}$, let $\vec{AE}=\lambda\vec{b}$
$\lambda\vec{b}.d\hat{k}=\lambda|\vec{b}|d\cos\theta$
for some real number $\lambda$ and $\hat{k}$ being unit normal vector in the direction $\vec{AB}$
$$\theta=\cos^{-1}\frac{\vec{b}.\hat{k}}{|\vec{b}|}$$
Now, we can also say $$\frac{\vec{b}.\vec{n}}{|\vec{b}||\vec{n}|}=\frac{\vec{b}.K\hat{n}}{K|\vec{b}|}=\frac{\vec{b}.\hat{n}}{|\vec{b}|}$$
Where $K$ is the magnitude of $\vec{n}$
That should mean from $1$:
$$\sin^{-1}\frac{\vec{b}.\hat{n}}{|\vec{b}|}=\cos^{-1}\frac{\vec{b}.\hat{k}}{|\vec{b}|}$$ (I realised that this result wouldn't have required this long derivation only after deriving it.)
Question $1$: Is the derivation so far correct?
Question $2$: How to express $$\cos^{-1}\frac{\vec{b}.\hat{k}}{|\vec{b}|}$$ as $$\sin^{-1}\frac{\vec{b}.\hat{n}}{|\vec{b}|}?$$( It is known that $\hat{k}$ is perpendicular to $\hat{n}$)
Because $\cos^2\theta+\sin^2\theta=1$ and $$ \cos\theta=\frac{\vec{b}\cdot\hat{\boldsymbol{k}}}{|\vec{b}|} $$ we only have to show that $$ \frac{(\vec{b}\cdot\hat{\boldsymbol{k}})^2}{|\vec{b}|^2}+\frac{(\vec{b}\cdot\hat{\boldsymbol{n}})^2}{|\vec{b}|^2}=1 $$ holds. This follows from the fact that $\hat{\boldsymbol{k}}$ and $\hat{\boldsymbol{n}}$ are orthogonal unit vectors by which the squared length of $\vec{b}$ is equal to the sum of the squares of the components of $\vec{b}$ given in the basis $\hat{\boldsymbol{k}}$ and $\hat{\boldsymbol{n}}\,.$ If you remember how to get those components from projections this means nothing else than $$ |\vec{b}|^2=(\vec{b}\cdot\hat{\boldsymbol{k}})^2+(\vec{b}\cdot\hat{\boldsymbol{n}})^2\,. $$