Suppose that we have a convex spherical quadrilateral, and we know its internal angles $\alpha,\beta,\gamma$ and $\delta$. Pairs of opposite sides of the quadrilateral are pieces of distinct great circles, and so may be continued to their antipodal meeting points. In this way we get two distinct convex spherical lunes (or bigons as some like to call them), none of whose vertices are vertices of the quadrilateral, but whose intersection is the quadrilateral. Call the angles of these two lunes $\theta$ and $\varphi$. Is it possible to determine these angles given only $\alpha,\beta,\gamma$ and $\delta$? A figure of the situation is shown below—we know the green angles and wish to find the red angles.
From the polar/dual law of cosines applied to the upper triangle,
$$\cos\varphi=-\cos(\pi-\alpha)\cos(\pi-\beta)+\sin(\pi-\alpha)\sin(\pi-\beta)\cos AB,$$
$\varphi$ determines the edge length $AB$ and vice-versa, assuming the angles $\alpha,\beta$ are given. So your question is whether the edge lengths can be determined from $\alpha,\beta,\gamma,\delta$.
(Equivalently by duality, your question is whether a spherical quadrilateral's angles can be determined from its edge lengths.)
The answer is no. Consider a "rectangle", defined by symmetry rather than right angles, so that $\alpha=\beta=\gamma=\delta$, $AB=CD$, $AD=BC$. Put the rectangle's centre at $(0,0,1)\in\mathbb S^2\subset\mathbb R^3$, and define the edges' outward normal vectors (tangent to the sphere):
$$e=(\cos\lambda,\;0,\;-\sin\lambda)$$
$$f=(0,\;\cos\mu,\;-\sin\mu)$$
$$g=(-\cos\lambda,\;0,\;-\sin\lambda)$$
$$h=(0,\;-\cos\mu,\;-\sin\mu).$$
We require
$$\cos(\pi-\alpha)=e\cdot f=f\cdot g=g\cdot h=h\cdot e$$
$$=\sin\lambda\sin\mu$$
to be constant, so either of $\lambda,\mu$ determines the other. We still have $1$ degree of freedom.
A vertex vector is perpendicular to the two edges' normal vectors; one of them is
$$\pm\frac{e\times f}{\lVert e\times f\rVert}=\pm\frac{(\sin\lambda\cos\mu,\;\cos\lambda\sin\mu,\;\cos\lambda\cos\mu)}{\sqrt{1-\sin^2\lambda\sin^2\mu}}$$
and an adjacent one is
$$\pm\frac{f\times g}{\lVert f\times g\rVert}=\pm\frac{(-\sin\lambda\cos\mu,\;\cos\lambda\sin\mu,\;\cos\lambda\cos\mu)}{\sqrt{1-\sin^2\lambda\sin^2\mu}}$$
so the cosine of the distance between these two vertices is
$$\pm\frac{e\times f}{\lVert e\times f\rVert}\cdot\frac{f\times g}{\lVert f\times g\rVert}=\pm\frac{1-2\sin^2\lambda+\sin^2\lambda\sin^2\mu}{1-\sin^2\lambda\sin^2\mu}$$
$$=\pm\frac{1+\cos^2\alpha-2\sin^2\lambda}{\sin^2\alpha}$$
which is not constant. In other words, the edge length is not determined by the angle $\alpha$.