Broken tree forming 30° with ground — original height A tree breaks so that its top touches the ground making a 30° angle with the ground. The distance from the root to the touching point is 10 m. What was the original height of the tree?
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A10√3
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B10/ √3
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C20 √3
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DNone of these
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E—
Answer
Correct Answer: 10√3
Explanation
Introduction / Context:When a tree breaks and its top touches the ground, the broken part forms a straight segment inclined to the ground. The original height equals the stump height plus the length of the broken part (since it was formerly vertical).
Given Data / Assumptions:
- Angle between broken top and ground = 30°.
- Horizontal distance from root to touching point d = 10 m.
- Let stump height be x and broken length be L; original height H = x + L.
Concept / Approach:The broken segment length L and its horizontal projection satisfy L cos 30° = d. The vertical drop from the break to ground equals x and is L sin 30°. Thus x = L/2, giving two equations to solve for L and x, then H.
Step-by-Step Solution:
L cos 30° = d ⇒ L (√3/2) = 10 ⇒ L = 20/√3x = L sin 30° = (20/√3) * (1/2) = 10/√3H = x + L = 10/√3 + 20/√3 = 30/√3 = 10√3Verification / Alternative check:Numerically, H ≈ 17.32 m, plausible for the geometry.
Why Other Options Are Wrong:10/√3 is just the stump; 20√3 is far too large given d = 10 m.
Common Pitfalls:Using H = L sin 30° only (omitting the stump), or taking d as the tree height.
Final Answer:10√3