A tree breaks and falls to the ground such that its upper part is still partially attached to its stem. At what height did it break, if the original height of the tree was 24 m and it makes an angle of 30° with the ground?

Correct option is D

Given:

The total height of the tree is 24 meters.

After the break, the upper part forms an angle of 30° with the ground. 

Formula Used: ​

​$$\sin 30^\circ = \frac{\text{perpendicular}}{\text{hypotenuse}}$$​​

Solution:

Let x be the height at which the tree broke.

The height of the remaining portion of the tree is 24−h24 - h24−x (Hypotenuse of right triangle)

The distance from the base of the tree to the point where the broken part touches the ground is the horizontal distance, which we'll denote as d. (Base of right triangle)

The height at which the tree broke is x (perpendicular of right triangle) 

Since we know the angle between the broken part and the ground is 30° , we can use the sine ratio:

​$$\text{Sin}(30^\circ) = \frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{x}{24 - x}$$  

using $$\text{Sin}30^\circ = \frac{1}{2}$$ 

$$\frac{{1}}{2} = \frac{x}{24 - x}$$ 

24 - x = 2x

3x = 24

x = $$\frac{24}{3} = 8$$ m

The tree broke at a height of 8 meters. 

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