Two identical right circular cones are resting on their curved surface as shown in the second row of the image above.
Let V be the top vertex and O (O') be the center of base circles and A(A') be the horizontal point on the circumference towards each other sides.
If the vertex of the cones are raised at a horizontal level of the base center, then the top and side view would look like in the first row image above.
Now, the cones are made to touch each other in the resting position of the second row.
It can be observed (experimentally/practically/DIY basis) that when the cones are touching each other such that their vertex as well as the curved surface too touch each other THEN THE BASE CIRCLE WILL TOUCH EACH OTHER at a point B(B')
1) Find the angle BOA (or B'O'A') in terms of the semi vertex angle of the cone?
2) Also find the elevation of point B(or B') over the vertex, there is a maximum value for this elevation, also find that!
Let a cone be resting on a plane, along generatrix $VB$ (see diagram below), and the perpendicular to the plane at $V$ meet line $BO$ at point $G$. From the similitude of triangles $GVO$ and $OVB$ we get: $$ OG={OV^2\over OB}. $$
Let now $GT$ be a line tangent to the base of the cone: the reflection of the cone about plane $GVT$ is tangent to the given cone at $T$. Hence the requested angle is $\angle AOT$ and: $$ \sin(\angle AOT)=\sin(\angle OGT)={OT\over OG}={OB\over OV^2/OB}= {OB^2\over OV^2}=\tan^2(\angle OVB). $$
EDIT.
Here's an animation made with GeoGebra, showing that the above formula is in perfect agreement with the angles computed by GeoGebra.