Question

three atoms with atomic radii of 4.0, 5.0, and 6.5 are arranged as in the figure. find the measure of the angle at the center of atom b inside the triangle. the measure of the angle is (square^{circ}). (do not round until the final answer. then round to the nearest degree as needed.)

Explanation:

Step1: Calculate the side lengths of the triangle

The length of side \(AC\) is \(4.0 + 6.5=10.5\), the length of side \(AB\) is \(6.5 + 5.0 = 11.5\), and the length of side \(BC\) is \(4.0+5.0 = 9.0\).

Step2: Use the Law of Cosines

The Law of Cosines formula is \(\cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}\), where \(a = AC = 10.5\), \(c=BC = 9.0\), and \(b = AB=11.5\).
Substitute the values: \(\cos B=\frac{10.5^{2}+9.0^{2}-11.5^{2}}{2\times10.5\times9.0}\)
First, calculate the squares: \(10.5^{2}=110.25\), \(9.0^{2} = 81\), \(11.5^{2}=132.25\)
Then, \(\cos B=\frac{110.25 + 81-132.25}{189}=\frac{191.25 - 132.25}{189}=\frac{59}{189}\approx0.312169\)

Step3: Find the angle \(B\)

Take the inverse cosine: \(B=\cos^{-1}(0.312169)\approx71.8^{\circ}\)

Answer:

\(72^{\circ}\)