1. **Problem statement:** A body of weight $w$ newton rests on a rough horizontal plane. A horizontal force $F$ newton acts on it trying to move it. The resultant reaction $R$ lies in the interval $]6,12]$ newton. We need to find the angle of friction $\theta$ in degrees. 2. **Relevant formula:** The resultant reaction $R$ is the vector sum of the normal reaction $N$ and the frictional force $f$. Since the body is on a horizontal plane, $N = w$ (weight) and friction $f = \mu N = \mu w$, where $\mu = \tan \theta$ is the coefficient of friction. 3. The resultant reaction $R$ magnitude is given by: $$ R = \sqrt{N^2 + f^2} = \sqrt{w^2 + (\mu w)^2} = w \sqrt{1 + \mu^2} = \frac{w}{\cos \theta} $$ 4. Given $R \in ]6,12]$, so: $$ 6 < R \leq 12 $$ Substitute $R = \frac{w}{\cos \theta}$: $$ 6 < \frac{w}{\cos \theta} \leq 12 $$ 5. Rearranging for $\cos \theta$: $$ \frac{w}{12} \leq \cos \theta < \frac{w}{6} $$ 6. Since $w$ is the weight, assume $w=10$ newton (typical for such problems) to find $\theta$ range: $$ \frac{10}{12} \leq \cos \theta < \frac{10}{6} $$ $$ 0.8333 \leq \cos \theta < 1.6667 $$ Since $\cos \theta$ cannot be more than 1, upper bound is 1. 7. So: $$ 0.8333 \leq \cos \theta \leq 1 $$ 8. Find $\theta$: $$ \theta = \cos^{-1}(0.8333) \approx 33.56^\circ $$ 9. Among the options (15°, 30°, 60°, 45°), the closest angle less than or equal to $33.56^\circ$ is $30^\circ$. **Final answer:** The angle of friction is approximately **30°**.