1. **Problem statement:** A 30 m ladder reaches a window 26 m high when placed at point A. After fixing the first window, the ladder is pushed back to point B, reducing the angle with the ground by half, still 30 m long, reaching a second window. Find the distance between points A and B. 2. **Known values:** Ladder length $L=30$ m, height to first window $h_1=26$ m. 3. **Step 1: Find the angle $\theta$ at point A.** Using the right triangle formed by the ladder, wall, and ground: $$\sin \theta = \frac{h_1}{L} = \frac{26}{30} = \frac{13}{15}$$ So, $$\theta = \arcsin\left(\frac{13}{15}\right)$$ 4. **Step 2: Find the horizontal distance from point A to the wall, call it $x_A$.** $$x_A = L \cos \theta = 30 \cos \theta$$ 5. **Step 3: After pushing the ladder back to point B, the angle with the ground is halved:** $$\theta_B = \frac{\theta}{2}$$ 6. **Step 4: Find the new horizontal distance $x_B$ from point B to the wall:** $$x_B = L \cos \theta_B = 30 \cos \left(\frac{\theta}{2}\right)$$ 7. **Step 5: The distance between points A and B is:** $$d = x_B - x_A = 30 \cos \left(\frac{\theta}{2}\right) - 30 \cos \theta = 30 \left(\cos \frac{\theta}{2} - \cos \theta\right)$$ 8. **Step 6: Calculate $\theta$ numerically:** $$\theta = \arcsin\left(\frac{13}{15}\right) \approx 1.1503 \text{ radians}$$ 9. **Step 7: Calculate $\cos \theta$ and $\cos \frac{\theta}{2}$:** $$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{13}{15}\right)^2} = \sqrt{1 - \frac{169}{225}} = \sqrt{\frac{56}{225}} = \frac{2\sqrt{14}}{15} \approx 0.4987$$ $$\cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}} = \sqrt{\frac{1 + 0.4987}{2}} = \sqrt{0.74935} \approx 0.8655$$ 10. **Step 8: Calculate $d$:** $$d = 30 (0.8655 - 0.4987) = 30 (0.3668) = 11.004$$ 11. **Step 9: Conclusion:** The distance between points A and B is approximately 11.0 m. 12. **Step 10: Choose the correct option:** $$d > 10.5 \text{ m}$$ **Final answer:** The distance between points A and B is approximately 11.0 m, which is greater than 10.5 m.