Question

three atoms with atomic radii of 4.0, 5.0, and 5.9 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: Find the side lengths of the triangle

The side lengths of the triangle are \(a = 4.0+5.9 = 9.9\), \(b=5.0 + 5.9=10.9\), \(c = 4.0+5.0 = 9.0\)

Step2: Use the Law of Cosines

The Law of Cosines formula is \(\cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}\)
Substitute \(a = 9.9\), \(b = 10.9\), \(c = 9.0\) into the formula:

$$
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Step3: Find the angle \(B\)

Since \(B=\cos^{-1}(0.338)\), then \(B\approx70.2^{\circ}\approx70^{\circ}\)

Answer:

\(70\)