The upper part of a tree breaks at a certain height and the broken top touches the ground at a point 10 m from the foot of the tree, making an angle of 60° with the ground. What was the original height of the tree?

Introduction / Context:
This is a classic broken tree problem that uses right triangle geometry and trigonometry. A tree breaks at some height above the ground, and the upper part falls but remains attached, touching the ground at a known distance from the foot of the tree. The broken part makes a given angle with the ground. Using this information, we must find the original full height of the tree before it broke.

Given Data / Assumptions:

• The upper part of the tree touches the ground at a point 10 m from the foot of the tree.
• The broken part makes an angle of 60 degrees with the horizontal ground.
• The tree was originally vertical and the ground is level.
• The tree broke at a height h above the ground.
• The broken top segment is straight and has length L.

Concept / Approach:
When the tree breaks, the portion above the break forms the hypotenuse of a right triangle with the vertical segment from the break point to the ground and the horizontal distance from the foot of the tree to the point where the top touches the ground. The angle between this broken part and the ground is 60 degrees. Using tan 60 = opposite / adjacent, we can find the height at which the tree broke. Then we use Pythagoras to find the length of the broken part and add both to get the original tree height.

Step-by-Step Solution:
Let h be the height of the break point above the ground. The horizontal distance from the foot of the tree to the point where the top touches the ground is 10 m. The angle between the broken part and the ground at that point is 60 degrees. In the right triangle formed, tan 60 = h / 10. So sqrt(3) = h / 10, giving h = 10√3 m. Now the length of the broken part L is the hypotenuse: L² = h² + 10² = (10√3)² + 10² = 300 + 100 = 400. Thus L = 20 m. Original height of the tree = h + L = 10√3 + 20 = 10(2 + √3) m.

Verification / Alternative check:
Check numerically: 10√3 is about 17.32 m, so the break point is around that height. The hypotenuse length 20 m is reasonable for a segment leaning at 60 degrees with a base of 10 m, since cos 60 = adjacent / hypotenuse = 10 / 20 = 0.5. This matches the standard value cos 60 = 0.5. Adding the two segments gives approximately 17.32 + 20 = 37.32 m, consistent with the exact expression 10(2 + √3).

Why Other Options Are Wrong:

• 20√3 m and 15√3 m: These represent other combinations of height and hypotenuse but do not match both tan 60 and Pythagoras simultaneously.
• 10√3 m: This is only the height up to the break point, not the complete original tree height.
• 10(2 - √3) m: This is much smaller and comes from using an incorrect sign in algebraic manipulation.

Common Pitfalls:
Students sometimes assume the broken part is equal to the original total height or misplace the 60 degree angle at the foot of the tree instead of at the ground contact of the broken top. Another frequent error is forgetting that the original height is the sum of the vertical part still standing plus the length of the broken segment. Drawing a clear diagram avoids most of these mistakes.

Final Answer:
The original height of the tree was 10(2 + √3) m.