1. Let the initial resultant be $R$.

Two forces have magnitudes: $P = 2F, \qquad Q = 3F$ and the angle between them is $\theta$.

So, by the law of cosines for vector addition, $R = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}$ Hence, $R^2 = (2F)^2 + (3F)^2 + 2(2F)(3F)\cos\theta$ $R^2 = 4F^2 + 9F^2 + 12F^2\cos\theta$ $R^2 = 13F^2 + 12F^2\cos\theta$

2. Now double force $Q$.

Then new force becomes: $Q' = 6F$ The new resultant is given to be doubled: $R' = 2R$

Now, $R'^2 = (2F)^2 + (6F)^2 + 2(2F)(6F)\cos\theta$ $R'^2 = 4F^2 + 36F^2 + 24F^2\cos\theta$ $R'^2 = 40F^2 + 24F^2\cos\theta$

But since $R' = 2R$, $R'^2 = 4R^2$ So, $40F^2 + 24F^2\cos\theta = 4\left(13F^2 + 12F^2\cos\theta\right)$

3. Solve for $\cos\theta$.

Expanding the right side: $40F^2 + 24F^2\cos\theta = 52F^2 + 48F^2\cos\theta$

Bring terms together: $40 - 52 = 48\cos\theta - 24\cos\theta$ $-12 = 24\cos\theta$ $\cos\theta = -\frac{1}{2}$

Therefore, $\theta = 120^\circ$

4. Check options.

• A: $90^\circ$ $\Rightarrow$ incorrect
• B: $60^\circ$ $\Rightarrow$ incorrect
• C: $30^\circ$ $\Rightarrow$ incorrect
• D: $120^\circ$ $\Rightarrow$ correct

Therefore, the correct answer is D.