How to get the coordinates of a point on a sphere when we only know the value of the side of a triangle? To elaborate a bit more on the title, I'm currently learning about spherical geometry and had a question regarding getting the coordinates of a point on a spherical triangle.
I have the following image:

Where $A$, $P$, and $Q$ are each points of a spherical triangle and $a$, $b$, and $c$ are their respective opposite sides.
The textbook that I'm using states that the coordinates for $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ are:
$$
\begin{align}
\overrightarrow{OP} & = (\sin{(b)},\ 0,\ \cos{(b)}) \\
\overrightarrow{OQ} & = (\sin{(c)}\cos{(A)},\ \sin{(c)}\sin{(A)},\ \cos{(c)})
\end{align}
$$
I'm wondering, how were these coordinates obtained? I initially thought that it was derived using the formula for the arc length (assuming the radius is $1$) but I can't seem to derive how the specific coordinates were obtained through that.
Any tips would be appreciated. Thanks!
Edit
To add some extra context, this arose when deriving the theorem for the law of cosines on a sphere.
Since $a = \angle POQ$, we can derive:
$$
\begin{align}
\cos{(a)} & = \left\langle \overrightarrow{OP}, \overrightarrow{OQ} \right\rangle \\
& = \langle (\sin{(b)},\ 0,\ \cos{(b)}), (\sin{(c)}\cos{(A)},\ \sin{(c)}\sin{(A)},\ \cos{(c)}) \rangle \\
& = \cos{(b)}\cos{(c)} + \sin{(b)}\sin{(c})\cos{(A)}
\end{align}
$$
 A: The coordinates of $P$ and $Q$ do not require any spherical trigonometry. They may be intended to help you acquire knowledge of spherical trigonometry, but where they come from is from ordinary trigonometry of triangles in planes, within the context of a three-dimensional Cartesian coordinate system.
So each point has an $x,$ $y,$ and $z$ coordinate in the Cartesian coordinate system.
For $P,$ consider the figure below.

In this figure, a right triangle is constructed with one side lying along
the line $OA$, which is the $z$ axis,
and the segment $OP$ as hypotenuse.
Saying that the arc from $A$ to $Q$ has angle $b$ is the same as saying that the angle of that triangle at vertex $O$ is $b.$
The length of the hypotenuse $OP$ is $1$
(since the radius of the sphere is $1$), 
so that means the lengths of the two legs
of the triangle are $\sin (b)$ and $\cos (b)$ as shown in the figure.
The side of length $\cos(b)$ lies along the $z$ axis, so that is the $z$ coordinate of $P$.
The side of length $\sin(b)$ is parallel to the $x$ axis, so that is the $x$ coordinate of $P$.
Since $P$ lies in the $x,z$ plane, its $y$ coordinate is zero.
Therefore the $(x,y,z)$ Cartesian coordinates of $P$ are
$(\sin(b), 0, \cos(b)).$
To get the $z$ coordinate of $Q$ we can construct the triangles shown in red below:

We construct a line segment from $Q$ that meets the $z$ axis at a right angle.
This is one side of a right triangle whose other side is the red segment along the $z$ axis. The angle of that triangle at vertex $O$ is $c.$ The hypotenuse $OQ$ is $1$, so that means the lengths of the two legs
of the triangle are $\sin (c)$ and $\cos (c)$ as shown in the figure.
In particular, the side along the $z$ axis, which is the $z$ coordinate of $Q$, is $\cos (c).$
The other red triangle in the figure is congruent to the first one; together they make a rectangle with sides $\cos c$ and $\sin c.$
For the $x$ and $y$ coordinates, consider the figure below.

In this figure, two sides of the rectangle from the first figure are repeated,
but a new right triangle has been constructed in the $x,y$ plane.
The hypotenuse of the triangle is one of the sides of the rectangle from the
first figure; its length is $\sin (c)$.
Since the angle at vertex $O$ in this triangle is $A,$
the two legs are gotten by multiplying the hypotenuse by the sine or cosine of $A.$
So we get legs of length
$\sin(c) \cos(A)$ and $\sin(c) \sin(A)$ as marked in the figure.
The side marked $\sin(c) \cos(A)$ lies along the $x$ axis, so its length is the $x$ coordinate of $Q$.
The side marked $\sin(c) \sin(A)$ is parallel to the $y$ axis, so its length is the $y$ coordinate of $Q.$
And there you have it:


*

*The $x$ coordinate of $Q$ is $\sin(c) \cos(A)$.

*The $y$ coordinate of $Q$ is $\sin(c) \sin(A)$.

*The $z$ coordinate of $Q$ (from before) is $\cos(c)$.


That is, the coordinates of $Q$ are $(\sin(c) \cos(A),\sin(c) \sin(A),\cos(c))$.
A: The distance $\overrightarrow{OP}=\overrightarrow{OQ}=1$. The sides of the triangle are then measured as angles since the length of an arc subtending angle $\theta$ is $s-r\theta$ and $r=1$. 
The $x$ co-ordinate of $P$ is $\overrightarrow{OP}\cos(\bar{b})=1$. $\cos(\frac{\pi}{2}-b)=\sin(b)$. $P$ lies in the $xz$-plane therefore its $y$ co-ordinate is $0$. The $z$ co-ordinate of $P$ is $\overrightarrow{OP}\sin(\bar{b})=1$. $\sin(\frac{\pi}{2}-b)=\cos(b)$.
The $z$ co-ordinate of $Q$ is $\overrightarrow{OQ}\sin(\bar{c})=1$. $\sin(\frac{\pi}{2}-c)=\cos(c)$. The projection of $Q$ onto the $xy$-plane is $R$ say. Then $\overrightarrow{OR}$ has length $\overrightarrow{OQ}\cos(\bar{c})=\sin(c)$. The $x$ and $y$ co-ordinates of $R$ are $\overrightarrow{OR}\cos(A)=\sin(c)\cos(A)$ and $\overrightarrow{OR}\sin(A)=\sin(c)\sin(A)$.
