Alright — let's do this. For an inclined plane, the relationship between the height ($h$) to which a load is raised, the length of the inclined plane ($L$, which is the distance moved by the effort), and the angle of inclination ($\theta$) is given by the sine function: $$\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{h}{L}$$ Step 1: Identify the given values. Angle of inclination, $\theta = 45^\circ$. Height the load is moved, $h = 20 \text{ m}$. Step 2: Rearrange the formula to solve for the distance moved by the effort ($L$). $$L = \frac{h}{\sin(\theta)}$$ Step 3: Substitute the given values into the formula. $$L = \frac{20 \text{ m}}{\sin(45^\circ)}$$ We know that $\sin(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.7071$. $$L = \frac{20 \text{ m}}{0.7071}$$ $$L \approx 28.284 \text{ m}$$ Step 4: Compare the calculated value with the given options. The calculated distance is approximately $28.28 \text{ m}$. This matches option (A). The final answer is $\boxed{\text{A) } 28.28 \text{ m}}$. Send me the next one 📸