1. **Problem statement:** A body of weight $w$ newton is 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. **Understanding the forces:** The weight $w$ acts vertically downward. The horizontal force $F$ acts horizontally. The resultant reaction $R$ is the vector sum of the normal reaction $N$ (vertical) and the frictional force $f$ (horizontal). 3. **Resultant reaction:** Since the body is on a horizontal plane, the normal reaction $N = w$ (weight). The frictional force $f = \mu N = \mu w$, where $\mu = \tan \theta$ is the coefficient of friction. 4. **Magnitude of resultant reaction:** $$ R = \sqrt{N^2 + f^2} = \sqrt{w^2 + (\mu w)^2} = w \sqrt{1 + \mu^2} = w \sec \theta $$ 5. **Given:** $$ 6 < R \leq 12 $$ 6. **Assuming $w = 6$ newton (since $R$ lower bound is 6), then:** $$ 6 < 6 \sec \theta \leq 12 $$ Divide all parts by 6: $$ 1 < \sec \theta \leq 2 $$ 7. **Recall:** $\sec \theta = \frac{1}{\cos \theta}$, so $$ 1 < \frac{1}{\cos \theta} \leq 2 \implies \frac{1}{2} \leq \cos \theta < 1 $$ 8. **Find $\theta$:** $$ \cos \theta \geq \frac{1}{2} \implies \theta \leq 60^\circ $$ and $$ \cos \theta < 1 \implies \theta > 0^\circ $$ 9. **Angle of friction $\theta$ must be one of the options:** 15°, 30°, 60°, 45°. Since $\theta \leq 60^\circ$ and $\theta > 0^\circ$, all options except possibly 60° fit. 10. **Check which angle satisfies $R \leq 12$:** For $\theta = 60^\circ$, $$ R = w \sec 60^\circ = 6 \times 2 = 12 $$ which is the upper bound. For $\theta = 45^\circ$, $$ R = 6 \sec 45^\circ = 6 \times \sqrt{2} \approx 8.49 $$ which is within the interval. For $\theta = 30^\circ$, $$ R = 6 \sec 30^\circ = 6 \times \frac{2}{\sqrt{3}} \approx 6.93 $$ which is also within the interval. For $\theta = 15^\circ$, $$ R = 6 \sec 15^\circ = 6 \times 1.0353 \approx 6.21 $$ which is also within the interval. 11. **Conclusion:** The angle of friction $\theta$ can be any of the given options except those that produce $R$ outside the interval. Since the problem asks for the measure of the angle of friction and the resultant reaction is in $]6,12]$, the maximum angle that satisfies this is $60^\circ$. **Final answer:** $\boxed{60^\circ}$