Discussion
A tree bent by the wind .The top of the tree meets the ground at an angle of 60°.If the distance between the top of the foot be 8 m then what was the height of the tree?
Answer: B
The bent tree splits into a standing stump \( h \) and a bent-over part \( L \), which together made up the original height. The 8 m is the distance from the foot to where the top touches the ground.
\[ \cos 60^{\circ} = \frac{8}{L} \quad\Longrightarrow\quad L = \frac{8}{1/2} = 16 \text{ m} \]
\[ \tan 60^{\circ} = \frac{h}{8} \quad\Longrightarrow\quad h = 8\sqrt{3} \text{ m} \]
\[ \text{original height} = h + L = 8\sqrt{3} + 16 = 8(\sqrt{3} + 2) \text{ m} \]
Numerically that is about 29.9 m. The stump and the bent part must be added - reporting \( 8\sqrt{3} \) alone gives only the part still upright.
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