A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Step-by-Step Solution
Key Concept: Model the situation as a right‑angled triangle. Let the stump height be \(x\) metres and the broken part length be \(L\) metres. The broken part makes an angle of \(30^{\circ}\) with the ground, so \(\cos30^{\circ}=\dfrac{8}{L}\) and \(\sin30^{\circ}=\dfrac{x}{L}\). Use the relation \(\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}\) and \(\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}\) together with the fact that the original height \(H = x + L\).
1. Draw a right‑angled triangle:\
- Base (ground) = 8 m (distance from foot of tree to point where top touches ground).\
- Hypotenuse = length of broken part = \(L\).\
- Height = stump height = \(x\).\
- Angle between hypotenuse and ground = \(30^{\circ}\).
2. Apply cosine definition:\
$$\cos 30^{\circ}=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{8}{L}$$\
Hence \(L = \frac{8}{\cos 30^{\circ}}\).
3. Apply sine definition:\
$$\sin 30^{\circ}=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{x}{L}$$\
So \(x = L\sin 30^{\circ}=\frac{L}{2}\).
4. Express original height:\
The original height of the tree \(H\) is the sum of the stump height and the broken part length:\
$$H = x + L = \frac{L}{2}+L = \frac{3L}{2}.$$
5. Compute \(L\):\
\(\cos 30^{\circ}=\frac{\sqrt{3}}{2}\). Therefore\
$$L = \frac{8}{\frac{\sqrt{3}}{2}} = \frac{16}{\sqrt{3}} = \frac{16\sqrt{3}}{3}\;\text{m}.$$
6. Find \(H\):\
$$H = \frac{3}{2}\times \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}} = 8\sqrt{3}\;\text{m}.$$
7. Result: The height of the tree is \(8\sqrt{3}\) metres (approximately \(13.9\) m).
Correct Answer: 8√3 m (≈ 13.9 m)