Let $\alpha=2\beta$ in triangle ABC. Prove that $b<a<2b$.

Solution:

Since $\alpha=2\beta$, then $\alpha>\beta$, and consequently, $a>b$, and $\beta$ must be acut angle. Inequality $a<2b$ now follow from sinus theorem for triangle

$\frac {\sin\alpha}{a}=\frac {\sin\beta}{b}$ which after using $\sin 2\beta=2\sin\beta\cos\beta$ means $\cos\beta=\frac {a}{2b}<1$, since $\beta$ is acut, and this give $a<2b$.

Solution:

Since $\alpha=2\beta$, then $\alpha>\beta$, and consequently, $a>b$, and $\beta$ must be acut angle. Inequality $a<2b$ now follow from sinus theorem for triangle

$\frac {\sin\alpha}{a}=\frac {\sin\beta}{b}$ which after using $\sin 2\beta=2\sin\beta\cos\beta$ means $\cos\beta=\frac {a}{2b}<1$, since $\beta$ is acut, and this give $a<2b$.

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