Tejas is standing at the top of a building and observes a car at an angle of depression of 30° as it approaches the base of the building at a uniform speed. 6 seconds later, the angle of depression increases to 60°, and at that moment, the car is 25 m away from the building.
Generated by claude-sonnet-4-6 · 2026-06-15 10:36 · grounding stimulus
Model Answer
Let AB = height of building, C and D be the two positions of the car (C nearer, D farther). CB = 25 m.
(i) Height of the building:
From △ABC: tan 60° = AB/BC
⟹ AB = 25√3 m
(ii) Distance between the two positions:
From △ABD: tan 30° = AB/BD
⟹ BD = AB/tan 30° = 25√3 × √3 = 75 m
Distance CD = BD − BC = 75 − 25 = 50 m
(iii) Total time to reach foot of building from D:
Speed of car = CD/time = 50/6 = 25/3 m/s
Total distance from D to B = 75 m
Total time = 75 ÷ (25/3) = 75 × 3/25 = 9 seconds
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Explanation
- Examiner expects correct tan ratio setup for both angles.
- Height comes directly from the nearer triangle (60°, 25 m).
- For part (iii), find speed from the 50 m in 6 s, then apply it to the full 75 m — many students forget to use the total distance DB, not just DC.