Chapter 9.6, Problem 48PS

Solution

To find: The distance between two cities.

The distance between cities B and C is $\approx 524$ miles.

Given data: An airplane flies at $55°$ east of north from city A to city B at distance 470 miles. Another plane flies at $7°$ north of east from city A to city C, at distance 890 miles.

Method/Formula used:

  

In triangle ABC ‘Law of Cosine’ states:

  $\cos \angle A=\frac{b^2+c^2-a^2}{2bc}$ .

Calculations:

The locations of city A , B , and C are shown in Fig. 1.

  

The location of cities forms a triangle whose angle $\angle A$ is $28°$ , length of side AB equals 470 miles, side AC equals 890 miles.

From, Law of Cosines for triangle ABC of cities,

  $\cos \angle \text{A}=\frac{\text{A}{\text{C}}^2+\text{A}{\text{B}}^2-\text{B}{\text{C}}^2}{2\text{×AC×AB}}$

Substitute $28°$ for $\angle \text{A}$ , 470 for AB, and 890 for AC,

  $\begin{array}{c}\cos 28°=\frac{{470}^2+{890}^2-\text{B}{\text{C}}^2}{2\cdot 470\cdot 890}\\ \cos 28°=\frac{1013000-\text{B}{\text{C}}^2}{836600}\\ \text{BC}=\sqrt{1013000-\left(836600\right)\cdot \left(0.8830\right)}\\ \text{BC}=523.72\end{array}$

Therefore, the distance between cities B and C is $\approx 524$ miles.