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(B) Step $1$: Use the Law of Cosines to find $\cos A$.$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{5^2 + 6^2 - 4^2}{2(5)(6)} = \frac{25 + 36 - 16}{60

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$2A$

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(B) Step $1$: Use the Law of Cosines to find $\cos A$.$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{5^2 + 6^2 - 4^2}{2(5)(6)} = \frac{25 + 36 - 16}{60} = \frac{45}{60} = \frac{3}{4}$.Step $2$: Use the Law of Cosines to find $\cos C$.$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{4^2 + 5^2 - 6^2}{2(4)(5)} = \frac{16 + 25 - 36}{40} = \frac{5}{40} = \frac{1}{8}$.Step $3$: Use the double angle identity $\cos 2A = 2\cos^2 A - 1$.$\cos 2A = 2(\frac{3}{4})^2 - 1 = 2(\frac{9}{16}) - 1 = \frac{9}{8} - 1 = \frac{1}{8}$.Step $4$: Since $\cos C = \frac{1}{8}$ and $\cos 2A = \frac{1}{8}$, we conclude that $C = 2A$.

In $\triangle ABC$, with usual notations, if $a = 4$, $b = 5$, and $c = 6$, then $\cos A$ and $\cos C$ are calculated using the Law of Cosines. Find the value of $\cos C$ and $\cos A$ to determine the relationship between $\angle C$ and $\angle A$.

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