In a triangle ABC, \angle A=2\theta, \angle B=\angle C=4\theta and \theta satisfies 4 \sin^{2}\theta+2 \sin \theta-1=0. What is the ratio of BC to AB?

  1. A. \sqrt{5}-1
  2. B. \frac{(\sqrt{5}-1)}{2} ✓
  3. C. \frac{(\sqrt{5}-1)}{4}
  4. D. 2(\sqrt{5}-1)

Correct Answer: B. \frac{(\sqrt{5}-1)}{2}

Explanation

The sum of angles is 2\theta + 4\theta + 4\theta = 180^{\circ} \implies 10\theta = 180^{\circ} \implies \theta = 18^{\circ}. Thus, A = 36^{\circ} and C = 72^{\circ}. By the Sine Rule, \frac{BC}{\sin A} = \frac{AB}{\sin C} \implies \frac{BC}{AB} = \frac{\sin 36^{\circ}}{\sin 72^{\circ}} = \frac{\sin 36^{\circ}}{2\sin 36^{\circ}\cos 36^{\circ}} = \frac{1}{2\cos 36^{\circ}}. Since \cos 36^{\circ} = \frac{\sqrt{5}+1}{4}, the ratio is \frac{1}{2(\frac{\sqrt{5}+1}{4})} = \frac{2}{\sqrt{5}+1} = \frac{2(\sqrt{5}-1)}{4} = \frac{\sqrt{5}-1}{2}.

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