The problem describes a situation where a tree has broken due to the wind, with the top of the broken part forming a right triangle with the ground. Let's break it down:

• The tree is leaning at a 30° angle with the ground.
• The distance from the base (root) of the tree to the point where the top touches the ground is 15 meters.
• We need to find the height of the broken part, which is essentially the length of the tree that has bent over.

Solution:

In this scenario, the broken part of the tree forms the hypotenuse of a right-angled triangle. The distance from the root of the tree to where the top touches the ground (15 m) is the base of the triangle, and the height is the vertical distance that the tree would have extended before breaking.

Using trigonometry, we know: $\cos(\theta) = \frac{\text{base}}{\text{hypotenuse}}$ Where:

• $\theta = 30^\circ$
• Base = 15 m
• Hypotenuse = height of the broken part (which we need to find)

Thus, the equation becomes: $\cos(30^\circ) = \frac{15}{\text{Height of the broken part}}$ Since $\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$, we have: $0.866 = \frac{15}{\text{Height of the broken part}}$ Now, solve for the height: $\text{Height of the broken part} = \frac{15}{0.866} \approx 17.32 \, \text{m}$

Thus, the height of the broken part of the tree is approximately 17.32 meters.

Would you like further details or have any additional questions?

Here are five related questions to extend the problem:

1. How would the answer change if the angle were 45° instead of 30°?
2. What if the distance to the top of the tree was 20 meters? How would the height change?
3. How can we calculate the vertical height the top of the tree is from the ground?
4. Can this problem be solved using sine instead of cosine?
5. What would happen if the tree fell at a different angle, say 60°?

Tip: Always check if a right triangle setup is involved when dealing with angles and distances to apply trigonometric ratios properly.