Given two cones $C_1, C_2$ centered at the vertex of a unit sphere and knowing only the angle $x$ between the axis of the cones, would it is be possible to find the two conical angles at the vertex? I presume there isnt sufficient information for this. So if in addition I know that the diameters of the base circles of the two cones are in the ratio $d_1:d_2 = p:q$ for some natural numbers $p,q$, can I express the angles of the two cone in terms of this ratio? The attached picture sort of captures what I wish to say : 
Edit 1: The bigger and smaller cones are tangent to each other along a ruling, to the left as seen in the picture.
Let the half-angle of each cone be $\theta$, $\varphi$ with $0 < \varphi < \theta < \pi/2$. Then from your diagram, we have $$\theta - \varphi = x.$$ Furthermore, assume $0 < d_1 < d_2 < 1$. Then $$\sin \theta = \frac{d_2}{2}, \quad \sin \varphi = \frac{d_1}{2}.$$ Consequently if $d_1/d_2 = p/q$, we have $$q \sin \varphi = p \sin \theta = p \sin (x + \varphi) = p \left(\sin x \cos \varphi + \cos x \sin \varphi\right),$$ hence $$0 = p \sin x \cos \varphi + (p \cos x - q) \sin \varphi$$ or $$\tan \varphi = \frac{p \sin x}{q - p \cos x}.$$ Therefore, $$\varphi = \tan^{-1} \frac{\sin x}{q/p - \cos x}$$ and $\theta = x + \varphi$.