1. **Problem statement:** We have three atoms A, B, and C with atomic radii 5.1, 4.0, and 2.0 respectively. They form a triangle by connecting their centers. We need to find the measure of the angle at atom B inside this triangle. 2. **Understanding the problem:** The triangle is formed by the centers of the atoms. The sides of the triangle correspond to the distances between centers of atoms. Since atoms are touching, the distance between two atoms equals the sum of their radii. 3. **Calculate side lengths:** - Side AB = radius of A + radius of B = 5.1 + 4.0 = 9.1 - Side BC = radius of B + radius of C = 4.0 + 2.0 = 6.0 - Side AC = radius of A + radius of C = 5.1 + 2.0 = 7.1 4. **Given angle:** The angle at atom C is 17°. 5. **Use Law of Cosines to find angle at B:** Label sides opposite to angles as usual: - Let angle at B be $\theta_B$. - Side AC opposite angle B is 7.1. - Sides AB and BC adjacent to angle B are 9.1 and 6.0. The Law of Cosines states: $$\cos(\theta_B) = \frac{AB^2 + BC^2 - AC^2}{2 \times AB \times BC}$$ Substitute values: $$\cos(\theta_B) = \frac{9.1^2 + 6.0^2 - 7.1^2}{2 \times 9.1 \times 6.0}$$ Calculate numerator: $$9.1^2 = 82.81, \quad 6.0^2 = 36, \quad 7.1^2 = 50.41$$ $$82.81 + 36 - 50.41 = 68.4$$ Calculate denominator: $$2 \times 9.1 \times 6.0 = 109.2$$ So: $$\cos(\theta_B) = \frac{68.4}{109.2} \approx 0.6264$$ 6. **Find angle $\theta_B$:** $$\theta_B = \cos^{-1}(0.6264) \approx 51.2^\circ$$ 7. **Final answer:** Rounded to nearest degree, $$\boxed{51^\circ}$$ Thus, the measure of the angle at atom B inside the triangle is approximately 51 degrees.