If the sides of a triangle are in G.P., and its largest angle is twice the smallest, then the common ratio `r` satisfies the inequality `0

Options

A. `0 lt r lt sqrt(2)`
B. `1 lt r lt sqrt(2)`
C. `1 lt r lt 2`
D. none of these

Explanation

To solve the problem, we need to analyze the conditions given for the triangle whose sides are in geometric progression (G.P.) and where the largest angle is twice the smallest angle. Let's denote the sides of the triangle as follows:

1. Let the sides of the triangle be \( a/r \), \( a \), and \( ar \), where \( a > 0 \) and \( r > 1 \) (since the sides are in G.P. and \( ar \) is the largest side).

2. The smallest angle corresponds to the side \( a/r \) and we denote it as \( \theta \). The largest angle corresponds to the side \( ar \) and is given as \( 2\theta \).

Now, we can apply the Law of Sines, which states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is constant. Thus, we can write:

\[ \frac{a/r}{\sin \theta} = \frac{ar}{\sin 2\theta} \]

3. From the above equation, we can express it as:

\[ \frac{a/r}{\sin \theta} = \frac{ar}{2 \sin \theta \cos \theta} \]

4. Simplifying this gives:

\[ \frac{1}{r} = \frac{r}{2 \cos \theta} \]

5. Cross-multiplying leads to:

\[ 2 \cos \theta = r^2 \]

6. Therefore, we can express \( r^2 \) in terms of \( \cos \theta \):

\[ r^2 = 2 \cos \theta \]

7. Since \( \cos \theta \) must be positive for \( \theta \) to be an angle in a triangle, we have:

\[ 0 < \cos \theta \leq 1 \]

8. This implies:

\[ 0 < r^2 < 2 \]

9. Taking the square root of the inequality gives:

\[ 0 < r < \sqrt{2} \]

10. However, since we established that \( r > 1 \) (as the sides are in G.P. and \( ar \) is the largest side), we can combine these inequalities:

\[ 1 < r < \sqrt{2} \]

Thus, the common ratio \( r \) satisfies the inequality \( 1 < r < \sqrt{2} \).

### Answer:
The correct option is **b. \( 1 < r < \sqrt{2} \)**.