Let $1<m<3$. In $\triangle ABC$, if $2b=(m+1)a$ and $\cos A=\frac{1}{2}\sqrt{\frac{(m-1)(m+3)}{m}}$, prove that there are two values to the third side, one of which is $m$ times the other.
$\frac{b}{a}=\frac{m+1}{2}>1$
We need to prove that $c$ has two values, one of which is $m$ times the other.
Applying cosine law,
$$c^2=a^2+b^2-2ab\cos C$$
I am stuck here as $\cos C$ is not given. Is there any other method possible for solving this question?
By the law of cosines, $$a^2=b^2+c^2-2bc\cos A=\left(\frac{(m+1)a}{2}\right)^2+c^2-2\cdot\frac{(m+1)a}{2}\cdot c\cdot \frac 12\sqrt{\frac{(m-1)(m+3)}{m}}$$ from which we have $$\frac{(m-1)(m+3)}{4}a^2=-c^2+\frac{(m+1)ac}{2}\sqrt{\frac{(m-1)(m+3)}{m}}$$ Now dividing the both sides by $a^2$ and letting $x=c/a$ give $$\frac{(m-1)(m+3)}{4}=-x^2+\frac{(m+1)x}{2}\sqrt{\frac{(m-1)(m+3)}{m}},$$ i.e. $$x^2-\frac{(m+1)}{2}x\sqrt{\frac{(m-1)(m+3)}{m}}+\frac{(m-1)(m+3)}{4}=0$$ and so $$\begin{align}\frac ca&=x\\&=\frac 12\left(\frac{(m+1)}{2}\sqrt{\frac{(m-1)(m+3)}{m}}\pm\frac{(m-1)}{2}\sqrt{\frac{(m-1)(m+3)}{m}}\right)\\&=\frac 14\sqrt{\frac{(m-1)(m+3)}{m}},\quad\frac m4\sqrt{\frac{(m-1)(m+3)}{m}}\end{align}$$ so, we have $$c=\frac a4\sqrt{\frac{(m-1)(m+3)}{m}},\quad\frac {ma}4\sqrt{\frac{(m-1)(m+3)}{m}}$$ from which the claim follows.