Sides of the triangles are in G.P. Question:-
The sides of a triangle are in G.P. and it's largest angle is twice the smallest one. Prove that the common ratio of the G.P. lies in the interval $(1,\sqrt{2})$

Attempt at a solution:-

First of all, I calculated the range of the values of the common ratio $r$ of the G.P. by solving the following three inequalities simultaneously:-
$$ar^2+ar\gt a \\$$ 
$$ar^2+a \gt ar \\$$ 
$$a+ar\gt ar^2 \\$$
The range of the common ratio comes out to be $$r \in \left[ \dfrac{-1+\sqrt{5}}{2},\dfrac{1+\sqrt{5}}{2} \right]$$
And as stated in the question $\angle A= \angle B$
So, by using the sine rule,
$$\dfrac{ar^2}{sin2B}=\dfrac{ar}{sin3B}=\dfrac{a}{sinB}$$
And, also taking the help of the cosine law, I arrived at $$\cos^2B(4r)-2\cos B-r=0$$
Then, as $D \ge 0$ for the solutions of $\cos B$ to be real. We get the codition $4+16r^2 \gt 0$, which is always true. So, I tried bounding the roots of the quadratic of $\cos B$.

Getting Stuck:-
I think that's where I am getting stuck, or if you could provide me with any other path for arriving at the, solution. 
 A: Now, as you said, $ar^2$ is opposite a bigger angle than $a$, so:
$$ar^2 > a \implies r^2 > 1 \implies r > 1$$
Note that since the triangle has the angle $2B$, we know that $2B < 180^\circ$, or that $B < 90^\circ$, so $0 < \cos B < 1$.
By Law of Sines, we have:
$$\frac{ar^2}{\sin 2B}=\frac{a}{\sin B}$$
Divide both sides by $a$:
$$\frac{r^2}{\sin 2B}=\frac{1}{\sin B}$$
Take cross-product:
$$r^2\sin B=\sin 2B$$
Use the double-angle identity:
$$r^2\sin B=2\sin B\cos B$$
Divide both sides by $2\sin B$:
$$\frac{r^2}{2}=\cos B$$
We can combine this with $0 < \cos B < 1$. What do we get from this inequality?
I'll let you take it from here. Good luck!
A: First of all, letting a = 1, we can simplify our calculation a lot.
Lastly, in simplifying $\sin 3B$, you need the formula $\sin 3x = 3\sin x – 4 \sin^3 x$. The proof goes like this:-

By sine law, we have $\dfrac {1}{\sin \theta} = \dfrac {r^2}{\sin 2\theta}$.
After simplifying, we get $r = \sqrt 2 \sqrt \cos \theta$
By sine law again, $\dfrac {r}{\sin 3\theta} = \dfrac {1}{\sin \theta}$
Using the formula mentioned, we get $r = 3 – 4\sin^2 \theta$
From which, we get $r_{max} = … = 3$ occurred when $\theta = 0$; and $ r_{min} = … = 0$ occurred when $\theta = 60^0$.
Therefore, $r_{max} = \sqrt 2 \sqrt {\cos (0)} = \sqrt 2$.
Similarly, $r_{min} = \sqrt 2 \sqrt {\cos (60^0)} = 1$.
