Question

If the sides of a triangle are in $GP$ and the largest angle is twice thesmallest angle then the common ratio, which is greater than $1$, lies in the interval

• A. $(1,\sqrt{3})$
• B. $(1,\sqrt{2})$
• C. $(1,\frac{\sqrt{5}+1}{2})$
• D. none of these

Answer 1

The correct option is B $(1,\sqrt{2})$

Let the sides of the triangle be $\frac{a}{r},a,ar$ with $a>0$ and $r>1$

Let $\alpha$ be the smallest angle so that the largest angle $2\alpha$.

Then $\alpha$ is opposite to the side $\frac{a}{r}$ and $2\alpha$ is opposite to $ar$

Applying the sine formula to the triangle, we get
$\frac{\frac{a}{r}}{\sin \alpha }=\frac{ar}{\sin 2\alpha }\Rightarrow \frac{\sin 2\alpha }{\sin \alpha }=\frac{{ar}^2}{a}=r^2$

$\Rightarrow r^2=\frac{2\sin \alpha \cos \alpha }{\sin \alpha }=2\cos \alpha <2\Rightarrow r<\sqrt{2}$

$\because r\in (1,\sqrt{2})$