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?

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:

Verification / 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