Question

If the sides of a triangle are in G.P and its larger angle is twice the smallest, then the common ratio $r$ satisfies the inequality

• A. $0<r<\sqrt{2}$
• B. $1<r<\sqrt{2}$
• C. $1<r<2$
• D. none of these

Answer 1

The correct option is B $1<r<\sqrt{2}$

Let the sides of the triangle be $\frac{a}{r},a,ar$ with $r>0$ and $r>1$.
Let $\alpha$ be the smallest angle, so that the largest angle is $2\alpha$.
Then $\alpha$ is opposite to the side $\frac{a}{r}$ and
$2\alpha$ is opposite to the side $ar$.
Now, applying sine rule; we get
$\frac{\frac{a}{r}}{\sin \alpha }=\frac{2r}{\sin 2\alpha }$
$\Rightarrow \frac{\sin 2\alpha }{\sin \alpha }=\frac{2r}{\frac{a}{r}}$
$\Rightarrow \frac{\sin 2\alpha }{\sin \alpha }=r^2$
$\Rightarrow 2\cos \alpha =r^2$
Since $\alpha \ne 0$, therefore, $2\cos \alpha <2$.
$\Rightarrow r^2<2$
$\Rightarrow r<\sqrt{2}$
Hence, $1<r<\sqrt{2}$.