Three atoms with atomic radii of $4.0,5.0$, and 6.4 are arranged as in the figure. Find the measure of the angle at the center of atom $B$ inside the triangle.

Step 1 ：We are given three atoms with atomic radii of 4.0, 5.0, and 6.4. These are arranged in a triangle and we are asked to find the measure of the angle at the center of atom B.

Step 2 ：We can use the Law of Cosines to solve this problem. The Law of Cosines states that for any triangle with sides of lengths a, b, and c and an angle γ between sides a and b, the following equation holds: \(c² = a² + b² - 2ab\cos(γ)\).

Step 3 ：In this case, a = 4.0, b = 5.0, and c = 6.4. We can rearrange the equation to solve for \(\cos(γ)\).

Step 4 ：Substituting the given values into the rearranged equation, we find that \(\cos(γ) = 0.000999999999999801\).

Step 5 ：We can then use the arccos function to find the measure of the angle γ. This gives us γ = 1.56979632662823 radians.

Step 6 ：Converting this to degrees, we find that the measure of the angle at the center of atom B is approximately 89.94 degrees.

Step 7 ：Final Answer: The measure of the angle at the center of atom B inside the triangle is approximately \(\boxed{89.94}\) degrees.