Question

The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 degrees. The tree stands 20 ft off the hill. How high is the top of the tree from the base of the hill?

Answer 1

Answer:

Approximately $101\; \rm ft$ (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

• Let $\rm O$ denote the observer.
• Let $\rm A$ denote the top of the tree.
• Let $\rm R$ denote the base of the tree.
• Let $\rm B$ denote the point where line $\rm AR$ (a vertical line) and the horizontal line going through $\rm O$ meets. $\angle \rm B\hat{A}R = 90^\circ$.

Angles:

• Angle of elevation of the base of the tree as it appears to the observer: $\angle \rm B\hat{O}R = 51^\circ$.
• Angle of elevation of the top of the tree as it appears to the observer: $\angle \rm B\hat{O}A = 57^\circ$.

Let the length of segment $\rm BR$ (vertical distance between the base of the tree and the base of the hill) be $x\; \rm ft$.

The question is asking for the length of segment $\rm AB$. Notice that the length of this segment is $\mathrm{AB} = (x + 20)\; \rm ft$.

The length of segment $\rm OB$ could be represented in two ways:

• In right triangle $\rm \triangle OBR$ as the side adjacent to $\angle \rm B\hat{O}R = 51^\circ$.
• In right triangle $\rm \triangle OBA$ as the side adjacent to $\angle \rm B\hat{O}A = 57^\circ$.

For example, in right triangle $\rm \triangle OBR$, the length of the side opposite to $\angle \rm B\hat{O}R = 51^\circ$ is segment $\rm BR$. The length of that segment is $x\; \rm ft$.

$\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}$.

Rearrange to find an expression for the length of $\rm OB$ (in $\rm ft$) in terms of $x$:

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}$.

Similarly, in right triangle $\rm \triangle OBA$:

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}$.

Equate the right-hand side of these two equations:

$0.810\, x \approx 0.649\, (x + 20)$.

Solve for $x$:

$x \approx 81\; \rm ft$.

Hence, the height of the top of this tree relative to the base of the hill would be $(x + 20)\; {\rm ft}\approx 101\; \rm ft$.