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(D) In $	riangle ABC$,sides $a, b, c$ are in $GP$,so $b^2 = ac$. Given $C - A = 60^{\circ}$. Using the Law of Cosines,$\cos A = \frac{b^2 + c^2 - a^2

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$\frac{\sqrt{13}-1}{4}$

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(D) In $	riangle ABC$,sides $a, b, c$ are in $GP$,so $b^2 = ac$. Given $C - A = 60^{\circ}$. Using the Law of Cosines,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ and $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$. Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$. Substituting $a = 2R \sin A, b = 2R \sin B, c = 2R \sin C$,we get $\sin^2 B = \sin A \sin C$. Using $2 \sin A \sin C = \cos(A - C) - \cos(A + C)$,we have $2 \sin^2 B = \cos(60^{\circ}) - \cos(180^{\circ} - B) = \frac{1}{2} + \cos B$. $2(1 - \cos^2 B) = \frac{1}{2} + \cos B$. $4 - 4 \cos^2 B = 1 + 2 \cos B$. $4 \cos^2 B + 2 \cos B - 3 = 0$. Solving for $\cos B$ using the quadratic formula: $\cos B = \frac{-2 \pm \sqrt{4 - 4(4)(-3)}}{2(4)} = \frac{-2 \pm \sqrt{52}}{8} = \frac{-2 \pm 2\sqrt{13}}{8} = \frac{-1 \pm \sqrt{13}}{4}$. Since $B$ is an angle in a triangle and $\cos B > 0$ is required for the progression to hold with the given conditions,$\cos B = \frac{\sqrt{13}-1}{4}$.

In $\triangle ABC$,if the sides $a, b, c$ are in geometric progression and the largest angle exceeds the smallest angle by $60^{\circ}$,then $\cos B$ is equal to

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