Question

An evergreen tree is supported by a wire extending from 1.5 feet below the top of the tree to a stake in the ground. The wire is 24 feet long and forms a 58° angle with the ground. how tall is the tree?​

Answer 1

Answer:

Approximately $21.9\; \rm ft$ (assumption: this tree is perpendicular to the ground.)

Step-by-step explanation:

Refer to the diagram attached (not drawn to scale.)

Label the following points:

• $\rm S$: stake in the ground.
• $\rm A$: top of the tree.
• $\rm B$: point where the wire is connected to the tree.
• $\rm C$: point where the tree meets the ground.

Segment $\rm SB$ would then denote the wire between the tree and the stake. The question states that the length of this segment would be $24\; \rm ft$. Segment $\rm AB$ would represent the $1.5\; \rm ft$ between the top of this tree and the point where the wire was connected to the tree.

The question is asking for the height of this tree. That would correspond to the length of segment $\rm AC$.

If this tree is perpendicular to the ground, then $\rm \angle A\hat{C}S =90^\circ$. Triangle $\rm \triangle BCS$ would be a right triangle with segment $\rm SB$ as the hypotenuse.

The question states that the angle between the wire (segment $\rm SB$) and the ground (line $\rm SC$) is $58^\circ$. Therefore, $\rm \angle A\hat{S}C = 58^\circ$.

Notice, that in right triangle $\rm \triangle BCS$, segment $\rm BC$ is the side opposite to the angle $\rm \angle B\hat{S}C = 58^\circ$. Therefore, the length of segment $\rm BC\!$ could be found from the length of the hypotenuse (segment $\rm SB$) and the cosine of angle $\rm \angle B\hat{S}C = 58^\circ\!$.

$\displaystyle \cos\left(\rm \angle B\hat{S}C\right) = \frac{\text{length of \mathrm{BC} }}{\text{length of \mathrm{SB} }} \quad \genfrac{}{}{0em}{}{\leftarrow\text{opposite}}{\leftarrow \text{hypotenuse}}$.

Rearrange to obtain:

$\begin{aligned}& \text{length of \mathrm{BC} } \\ &= (\text{length of \mathrm{SB} }) \cdot \cos\left(\angle \mathrm{B\hat{S}C}\right)\\ &= \left(24\; \rm ft\right) \cdot \cos\left(58^\circ\right) \approx 20.35\; \rm ft\end{aligned}$.

In other words, the wire is connected to the tree at approximately $20.3\; \rm ft$ above the ground.

Combine that with the length of segment $\rm AB$ to find the height of the entire tree:

$\begin{aligned}&\text{height of the tree} \\ &= \text{length of \mathrm{AC} } \\ &= \text{length of \mathrm{AB} } + \text{length of \mathrm{BC} }\\ &\approx 20.35\; \rm ft + 1.5\; \rm ft \\ &\approx 21.9\; \rm ft\end{aligned}$.