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 $$ \approx 109.47^{0}. $$
Let each edge of regular tetrahedron be $a$ then the distance of each vertex from the center is $$=\frac{a}{2}\sqrt{\frac{3}{2}}$$ (For derivation see HCR's Formula for platonic solids)
Now, draw perpendicular to one of the edges of tetrahedron to get a right triangle, now the angle subtended by each edge at the center of regular tetrahedron $$=2\sin^{-1}\left(\frac{\frac{a}{2}}{\frac{a}{2}\sqrt{\frac{3}{2}}}\right)=2\sin^{-1}\left(\sqrt{\frac{2}{3}}\right)\approx109.4712206^{o}$$
Your description is slightly unusual, but what you have is a regular tetrahedron, and from the Wikipedia entry here, you can take the centre to be the origin and two of the vertices to be $$(0, 1, 1/\sqrt{2}), \quad (0, -1, 1/\sqrt{2})$$
This gives you an isosceles triangle with sides $\sqrt{\frac32}, \sqrt{\frac32}, 2$ and you can use the cosine rule from elementary trigonometry ($c^2 = a^2 + b^2 -2ab\cos\theta$) to get $$4 = \frac32 + \frac32 - 2.\frac32\cos\theta$$
so that
$$\cos\theta = -\frac13 \text{ and finally, we get } \theta = \cos^{-1}\left(-\frac13\right)=\pi-\cos^{-1}\left(\frac13\right)\approx109.4712206^{o}$$
