I'm studying spherical geometry and had a question regarding triangles on spheres.
The book that I'm using says that if we have triangle $\triangle ABC$ and we want to obtain the dual triangle $\triangle^* ABC$, then we do:
$$ \begin{align} A^* & = \dfrac{B \times C}{\sin{(a)}} \\ B^* & = \dfrac{A \times C}{\sin{(b)}} \\ C^* & = \dfrac{A \times B}{\sin{(c)}} \end{align} $$
What I'm wondering is why do we divide by sine here?
I know that the cross product between two vectors is:
$$ \mathbf{a} \times \mathbf{b} = \Vert \mathbf{a} \Vert \Vert \mathbf{b} \Vert \sin{(\theta)} \mathbf{n} $$
and so it seems that dividing by each respective sine would leave us with
$$ \mathbf{a} \times \mathbf{b} = \Vert \mathbf{a} \Vert \Vert \mathbf{b} \Vert \mathbf{n} $$
I'm just confused as to what the division accomplishes. My intuition of how the sine in the cross product measures "how perpendicular" the two vectors are tells me that it has something to do with normalization, but I'm not 100% sure on that.
Any tips are appreciated.
This is really just an elaboration of @ancientmathematician's comment. From your cross product formula, since $B,C$ are vectors with length $1$, one obtains that
$$||A^*|| = \frac{1}{|\sin(a)|} \cdot ||B||\cdot ||C|| \cdot \sin(\sphericalangle(B,C)).$$
Now, since the distance function $d_S$ on the sphere $S = S^2$ is defined by
$$d_S(X,Y) := \arccos(\langle X,Y\rangle)$$ and since $\cos(\sphericalangle(B,C)) = \frac{\langle B,C \rangle}{||B||\cdot ||C||} = \langle B,C\rangle$, we obtain that $a := d_S(B,C) = \sphericalangle (B,C)$. (Here, $\langle \cdot , \cdot\rangle$ denotes the standard inner product on $\mathbb R^3$.)
Combining everything, we obtain $||A^*||= \frac{\sin(a)}{|\sin(a)|} = 1$, since $a \in (0,\pi)$ as it is the length of a side of a spherical triangle.
The argument is of course the same for $B^*$ and $C^*$.