Chapter 6.2, Problem 54E

Determining a Distance A woman standing on a hill sees a flagpole that she knows is 60 ft tall. The angle of depression to the bottom of the pole is 14°, and the angle of elevation to the top of the pole is 18°. Find her distance x from the pole.

To determine

To Find: The distance $x$ between her and the pole.

Answer to Problem 54E

The distance between her and pole is $104.4932\text{ft}$.

Explanation of Solution

Given:

The height of flagpole is $60\text{ft}\text{.}$

Angle of depression from the bottom of the pole is $14°$.

Angle of elevation from the top of the pole is $18°$.

Calculation:

The below Figure show the angle between women to base of the flag and women to top of the flag.

Figure (1)

Let $y$ is height of $\text{CD}$ and $\left(60-y\right)$ is the height of $\text{BC}$ as shown in the above figure.

Formula Used: Trigonometry property of tangent..

$\text{tan}\theta =\left(\frac{\text{opposite}}{\text{adjacent}}\right)$

Consider the bottom triangle $\left(\text{ACD}\right)$

Use tangent property of trigonometry to calculate the value of $y$..

Substitute $14°$ for $\theta$, $y$ for opposite and $x$ for adjacent.

$\begin{array}{c}\text{tan14}°=\left(\frac{y}{x}\right)\\ x=\left(\frac{y}{\tan 14°}\right)\\ y=\left(0.2493\times x\right)\left(\because \tan 14°=0.2493\right)\\ y=0.2493x\end{array}$ (I)

Consider the top triangle $\left(ABC\right)$

Use tangent property of trigonometry to calculate the value of $x$..

Substitute $18°$ for $\theta$, $y$ for opposite.

$\begin{array}{c}\text{tan18}°=\left(\frac{60-y}{x}\right)\\ x=\left(\frac{60-y}{\tan 18°}\right)\\ \left(60-y\right)=\left(x\times \text{tan}18°\right)\end{array}$ (II)

Substitute the value of $y$ from equation (I) to equation (II).

$\begin{array}{c}y=0.2493x\\ \left(60-y\right)=\left(x\times \text{tan}18°\right)\\ \left(60-0.2493x\right)=\left(x\times \text{tan}18°\right)\end{array}$

Take $\left(0.2493x\right)$ at the right hand side to simplify the equation.

$\begin{array}{c}x\left(\text{tan}18°+0.2493\right)=60\\ x\left(0.3249+0.2493\right)=60\left(\because \text{tan}18°=0.3249\right)\\ x\left(0.5742\right)=60\end{array}$

Take (0.5742) at right hand side to calculate the value of $x$

$\begin{array}{l}x=\left(\frac{60}{0.5742}\right)\\ x=104.4932\text{ft}\end{array}$

Therefore, the distance between her and pole is $104.4932\text{ft}$.