Chapter 6, Problem 1P

To determine

To find: The distance between the church and the City Hall in the given map.

Answer to Problem 1P

The distance between the church and the City Hall is $1.41\text{mi}$.

Explanation of Solution

Given:

The map is given below as,

Figure (1)

Calculation:

The figure below shows the labeling of the map between church and the city hall, where the church is at point $\text{A}$ and city hall is at point $\text{B}$ and the distance between them is $x$.

Figure (2)

In the triangle $\Delta \text{AOB}$.

$\begin{array}{c}\angle \text{AOB}+30°+25°=180°\left(\text{angle}\text{sum}\text{property}\right)\\ \angle \text{AOB}=180°-55°\\ =125°\end{array}$

Use Law of Sines in $\Delta \text{AOB}$.

$\frac{\sin \left(\angle \text{AOB}\right)}{\text{AB}}=\frac{\sin \left(\angle \text{OAB}\right)}{\text{BO}}$

Substitute $125°$ for $\angle \text{AOB}$, $30°$ for $\angle \text{OAB}$, $0.86$ for $\text{BO}$ and $x$ for $\text{AB}$ in the above equation.

$\begin{array}{c}\frac{\sin \left(125°\right)}{x}=\frac{\sin \left(30°\right)}{0.86}\\ \frac{0.819152}{x}=\frac{0.5}{0.86}\left(\sin 125°=0.819152\text{and}\sin 30°=0.5\right)\\ x\approx 1.41\end{array}$

Thus the distance between the church and the cirty hall is $1.41\text{mi}$.