Triangle Solving – Problem 27: Find the range of

In $\triangle ABC$, let the sides opposite $A,B,C$ be $a,b,c$. Suppose $5a^{2}+3b^{2}=3c^{2}$. Find the range of $\sin A$.

Step 1. From $5a^{2}+3b^{2}=3c^{2}$ we have $a^{2}=\frac{3}{5}(c^{2}-b^{2})$, so $c>b>0$.

Step 2. By the Law of Cosines, $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}$. Substituting yields $$\cos A=\frac{b^{2}+c^{2}-\frac{3}{5}(c^{2}-b^{2})}{2bc}=\frac{c^{2}+4b^{2}}{5bc}=\frac{1}{5}\left(\frac{c}{b}+\frac{4b}{c}\right).$$

Step 3. Let $t=\frac{c}{b}>1$. Then $\cos A=\frac{t^{2}+4}{5t}=\frac{t+\frac{4}{t}}{5}$.

Step 4. Also $a=b\sqrt{\frac{3}{5}(t^{2}-1)}$. The triangle inequality $a+b>c$ gives $$\sqrt{\frac{3}{5}(t^{2}-1)}+1>t\Rightarrow 1<t<4.$$

Step 5. On $t\in(1,4)$, we have $t+\frac{4}{t}\ge 4$ (equality at $t=2$), so $\cos A\ge\frac{4}{5}$.

Step 6. Therefore $\sin A=\sqrt{1-\cos^{2}A}\le\sqrt{1-\frac{16}{25}}=\frac{3}{5}$.

Step 7. Equality holds at $t=2$ (i.e. $c=2b$). As $t\to 1^{+}$ or $t\to 4^{-}$, $\cos A\to 1$ and $\sin A\to 0^{+}$.

Step 8. Hence $\sin A\in\left(0,\frac{3}{5}\right]$.

$$(0,\frac{3}{5}]$$