Solving Triangles - Part 1

Since the range of the function is in terms of the triangle's angles, we can set $c = 1$ to make calculations easier. Then $b^2 = a^2 + a$ .

As an acute triangle, $a^2 + b^2 > c^2$ , $b^2 + c^2 > a^2$ , and $a^2 + c^2 > b^2$ . Using $c = 1$ and $b^2 = a^2 + a$ , gives $a^2 + a^2 + a > 1$ or $a > \frac{1}{2}$ , $a^2 + a + 1 > a^2$ or $a > -1$ , and $a^2 + 1 > a^2 + a$ or $a < 1$ . Therefore, $\frac{1}{2} < a < 1$ .

By law of cosines, $\cos B = \frac{a^2 + b^2 - c^2}{2ac}$ , and since $b^2 = a^2 + a$ and $c = 1$ this simplifies to $\cos B = \frac{1 - a}{2a}$ . Similarly, $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ which simplifies to $\cos A = \frac{\sqrt{a + 1}}{2 \sqrt{a}}$ .

Using a trigonometric identity, $\sin^2 A = 1 - \cos^2 A = 1 - (\frac{\sqrt{a + 1}}{2 \sqrt{a}})^2 = \frac{3a - 1}{4a}$ and $\sin^2 B = 1 - \cos^2 B = 1 - (\frac{1 - a}{2a})^2 = \frac{(3a - 1)(a + 1)}{4a^2}$ .

Plugging these sine and cosine values in the given function $\frac{\sin^2 A}{\sin(B - A)} = \frac{\sin^2 A}{\sin B \cos A - \cos B \sin A}$ gives $\frac{\sqrt{3a - 1}}{2 \sqrt{a}}$ , which increases from $\frac{1}{2} < a < 1$ . Therefore, the range is $(\frac{\sqrt{3 \cdot \frac{1}{2} - 1}}{2 \sqrt{\frac{1}{2}}}, \frac{\sqrt{3 \cdot 1 - 1}}{2 \sqrt{1}})$ or $\boxed{(\frac{1}{2}, \frac{\sqrt{2}}{2})}$ .

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Edit (Thanks to the author's hint in comments below):

Note that $\cos 2A = 2 \cos^2 A - 1 = 2(\frac{\sqrt{a + 1}}{2 \sqrt{a}})^2 - 1 = \frac{1 - a}{2a} = \cos B$ . Therefore, $B = 2A$ , which makes the given expression $\frac{\sin^2 A}{\sin(B - A)} = \sin A$ . Therefore to be an acute triangle, $30° < \angle A < 45°$ , so the range is $(\sin 30°, \sin 45°)$ or $\boxed{(\frac{1}{2}, \frac{\sqrt{2}}{2})}$ .