1. **Problem statement:** Alice observes an object at the top of a tree 4 meters away from the tree base (Point A). She then moves 2 meters closer (Point B), and the tree height is 6 meters. We need to find angles at points A and B to the top of the tree, name these angles, compare their sizes, and find the distances from the object to points A and B. 2. **Define variables and picture:** - Tree height $h = 6$ m - Distance from Point A to tree base $d_A = 4$ m - Distance from Point B to tree base $d_B = 4 - 2 = 2$ m 3. **(i) Find angles at points A and B relative to top of tree:** These are angles of elevation — angles formed by looking up from ground points A and B to the top of the tree. Angle at A, $\theta_A = \tan^{-1}(\frac{h}{d_A}) = \tan^{-1}\left(\frac{6}{4}\right)$ Angle at B, $\theta_B = \tan^{-1}(\frac{h}{d_B}) = \tan^{-1}\left(\frac{6}{2}\right)$ Calculate exact values: $$ \theta_A = \tan^{-1}(1.5) \approx 56.31^\circ $$ $$ \theta_B = \tan^{-1}(3) \approx 71.57^\circ $$ 4. **(ii) Compare angles $\theta_A$ and $\theta_B$:** Since $71.57^\circ > 56.31^\circ$, angle B is larger than angle A. _Conclusion:_ The closer you are to the object, the larger the angle of elevation you observe. 5. **(iii) Find distances from A and B to the object at the top:** Distances are hypotenuses of right triangles formed by height and horizontal distances. Distance from A to object: $$ D_A = \sqrt{h^2 + d_A^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} \approx 7.21 \text{ m} $$ Distance from B to object: $$ D_B = \sqrt{h^2 + d_B^2} = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} \approx 6.32 \text{ m} $$