Problem 1

Prove that there is one and only one triangle whose side lengths are consecutive integers, and one of whose angles is twice as large as another.

Step 2 of 6: Law of Sines gives cos α = a/(2b)

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In plain words

This relation is a clean bridge between the angle condition $\angle A = 2\angle B$ and the side lengths: it says the cosine of the smaller angle is completely determined by the ratio of two of the sides, with no other unknowns involved.

\frac{b}{\sin\alpha}=\frac{a}{\sin 2\alpha} \;\Longrightarrow\; \cos\alpha=\frac{a}{2b}

Detailed analysis

By the Law of Sines, $\dfrac{b}{\sin\alpha}=\dfrac{a}{\sin2\alpha}$ . Since $\sin2\alpha=2\sin\alpha\cos\alpha$ , this simplifies to $\dfrac{\sin2\alpha}{\sin\alpha}=2\cos\alpha=\dfrac{a}{b}$ , hence $\cos\alpha=\dfrac{a}{2b}$ .