what will be the angle at the centre? Taken a tetrahedron of same edges, a point is taken inside it which is equidistant from all $4$ vertices, i.e if a sphere is made taking it as a centre, all the vertices will be on the sphere, now taking any two vertex and that centre (on the plane passing through that centre and these two vertices),could any one tell me what will be the angle at the centre? I could not draw here :(

The physics way I believe is the easier, above geometric methods.

Let four equal forces $F$ act on the center with tetrahedral symmetry. Taking equilibrium of forces along a center-vertex line and its opposite extension:

$$ 3 F \cos \alpha + F = 0; \;\alpha = \cos^{-1} \frac {1}3. $$

where we took $\alpha$ as the acute angle. Required angle is its supplement $$ \approx 109.47^{0}. $$