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

Introduction / Context:
This problem is a classic height-and-distance application involving a broken tree. The tree originally stood vertically, then broke at some height, and its top touched the ground forming a right triangle with the tree trunk and the ground. We use trigonometric ratios to determine the original height of the tree.

Given Data / Assumptions:

• The tree was originally vertical.
• The top portion broke and now touches the ground.
• The broken top makes an angle of 60° with the ground.
• The horizontal distance from the tree's foot to where the top touches the ground is 10 m.
• We ignore the thickness of the tree and assume it is a straight line segment.

Concept / Approach:
Once the tree breaks, the upper part becomes the hypotenuse of a right triangle whose base is the distance from the tree foot to the point where the top touches the ground. The angle between this hypotenuse and the ground is 60°. The vertical segment from the ground to the breaking point is one leg of the triangle, and the hypotenuse is the length of the broken part. The original height equals the sum of the stump height plus the length of the broken part (since that broken piece used to be vertical as well).

Step-by-Step Solution:
Let L be the length of the broken top (hypotenuse).Base (distance along ground) = 10 m.Angle with ground = 60°.cos(60°) = base / hypotenuse = 10 / Lcos(60°) = 1 / 2, so 1 / 2 = 10 / LTherefore, L = 10 * 2 = 20 m.Height of breaking point above ground = L * sin(60°)sin(60°) = √3 / 2, so stump height = 20 * (√3 / 2) = 10√3 m.Original height of tree = stump height + broken part length = 10√3 + 20.Factorizing: 10√3 + 20 = 10(√3 + 2) = 10(2 + √3) m.

Verification / Alternative check:
We can verify by reconstructing: if original height is 10(2 + √3), then the part above the breaking point is 20 m. Placing a 20 m segment at 60° with the ground gives a horizontal projection of 20 * cos(60°) = 10 m, which matches the given distance. The vertical projection 20 * sin(60°) equals 10√3 m, which is the stump height. Everything is consistent.

Why Other Options Are Wrong:
10√3 m: This is only the stump height, not the total height.20√3 m: Overestimates and does not match the trigonometric relationships.10(2 - √3) m: This is much smaller and would imply an unrealistic geometry.15(2 + √3) m: Arbitrary multiple that does not satisfy the given 10 m base condition.

Common Pitfalls:
Many learners forget that the original height is the stump plus the broken portion, not just one of them. Another frequent error is using tan instead of cos for the base, or misusing √3/2 and 1/2 for 60°. Carefully identifying which side is adjacent, opposite, and hypotenuse relative to the given angle is essential.

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