Is my derivation of the tetrahedral bond angle correct?

Question

Let $T$ be a regular tetrahedron with edge length $x$.

Let $A$ be one of the faces of $T$.

Let $P$ be the plane containing $A$.

Let $L$ be the line segment from the center of $A$ to one of the vertices of $A$.

Let $H$ be the shortest line segment from $P$ to the vertex of $T$ that is not in the plane $P$.

Let $E$ be the edge of $T$ that forms a right-angled triangle together with $L$ and $H$.

Since $A$ is by definition an equilateral triangle, then the length of $L$ is $\frac{x}{√3}$.

Therefore, the angle between $H$ and $E$ is $\sin^{-1}(\frac{1}{√3})$.

Let $\alpha = \sin^{-1}(\frac{1}{√3})$.

Since the tetrahedral bond angle is the remaining angle of an isosceles triangle whose other two angles are each equal to $\alpha$, then the tetrahedral bond angle is $$π-2\alpha=2(π/2-\alpha)=2(π/2-\arcsin(\frac{1}{√3}))=2\cos^{-1}(\frac{1}{√3}).$$

Answer 1

Yes, this looks OK to me. Well done!