Solving Applied Problems Using the Law of Cosines

Just as the Law of Sines provided the appropriate equations to solve a number of applications, the Law of Cosines is applicable to situations in which the given data fits the cosine models. We may see these in the fields of navigation, surveying, astronomy, and geometry, just to name a few.

Example 3: Using the Law of Cosines to Solve a Communication Problem

On many cell phones with GPS, an approximate location can be given before the GPS signal is received. This is accomplished through a process called triangulation, which works by using the distances from two known points. Suppose there are two cell phone towers within range of a cell phone. The two towers are located 6000 feet apart along a straight highway, running east to west, and the cell phone is north of the highway. Based on the signal delay, it can be determined that the signal is 5050 feet from the first tower and 2420 feet from the second tower. Determine the position of the cell phone north and east of the first tower, and determine how far it is from the highway.

Solution

For simplicity, we start by drawing a diagram similar to Figure 6 and labeling our given information.

A triangle formed between the two cell phone towers located on am east to west highway and the cellphone between and north of them. The side between the two towers is 6000 feet, the side between the left tower and the phone is 5050 feet, and the side between the right tower and the phone is 2420 feet. The angle between the 5050 and 6000 feet sides is labeled theta.

Figure 6

Using the Law of Cosines, we can solve for the angle [latex]\theta[/latex]. Remember that the Law of Cosines uses the square of one side to find the cosine of the opposite angle. For this example, let [latex]a=2420,b=5050[/latex], and [latex]c=6000[/latex]. Thus, [latex]\theta[/latex] corresponds to the opposite side [latex]a=2420[/latex].

[latex]\begin{array}{l}\begin{array}{l}\hfill \\ \text{ }{a}^{2}={b}^{2}+{c}^{2}-2bc\cos \theta \hfill \end{array}\hfill \\ \text{ }{\left(2420\right)}^{2}={\left(5050\right)}^{2}+{\left(6000\right)}^{2}-2\left(5050\right)\left(6000\right)\cos \theta \hfill \\ {\left(2420\right)}^{2}-{\left(5050\right)}^{2}-{\left(6000\right)}^{2}=-2\left(5050\right)\left(6000\right)\cos \theta \hfill \\ \text{ }\frac{{\left(2420\right)}^{2}-{\left(5050\right)}^{2}-{\left(6000\right)}^{2}}{-2\left(5050\right)\left(6000\right)}=\cos \theta \hfill \\ \text{ }\cos \theta \approx 0.9183\hfill \\ \text{ }\theta \approx {\cos }^{-1}\left(0.9183\right)\hfill \\ \text{ }\theta \approx 23.3^\circ \hfill \end{array}[/latex]

To answer the questions about the phone’s position north and east of the tower, and the distance to the highway, drop a perpendicular from the position of the cell phone, as in Figure 7. This forms two right triangles, although we only need the right triangle that includes the first tower for this problem.

The triangle between the phone, the left tower, and a point between the phone and the highway between the towers. The side between the phone and the highway is perpendicular to the highway and is y feet. The highway side is x feet. The angle at the tower, previously labeled theta, is 23.3 degrees.

Figure 7

Using the angle [latex]\theta =23.3^\circ[/latex] and the basic trigonometric identities, we can find the solutions. Thus

[latex]\begin{array}{l}\begin{array}{l}\hfill \\ \cos \left(23.3^\circ \right)=\frac{x}{5050}\hfill \end{array}\hfill \\ \text{ }x=5050\cos \left(23.3^\circ \right)\hfill \\ \text{ }x\approx 4638.15\text{feet}\hfill \\ \text{ }\sin \left(23.3^\circ \right)=\frac{y}{5050}\hfill \\ \text{ }y=5050\sin \left(23.3^\circ \right)\hfill \\ \text{ }y\approx 1997.5\text{feet}\hfill \\ \hfill \end{array}[/latex]

The cell phone is approximately 4638 feet east and 1998 feet north of the first tower, and 1998 feet from the highway.

Example 4: Calculating Distance Traveled Using a SAS Triangle

Returning to our problem at the beginning of this section, suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat? The diagram is repeated here in Figure 8.

A triangle whose vertices are the boat, the port, and the turning point of the boat. The side between the port and the turning point is 10 mi, and the side between the turning point and the boat is 8 miles. The side between the port and the turning point is extended in a straight dotted line. The angle between the dotted line and the 8 mile side is 20 degrees.

Figure 8

Solution

The boat turned 20 degrees, so the obtuse angle of the non-right triangle is the supplemental angle, [latex]180^\circ -20^\circ =160^\circ[/latex]. With this, we can utilize the Law of Cosines to find the missing side of the obtuse triangle—the distance of the boat to the port.

[latex]\begin{array}{l}{x}^{2}={8}^{2}+{10}^{2}-2\left(8\right)\left(10\right)\cos \left(160^\circ \right)\hfill \\ {x}^{2}=314.35\hfill \\ x=\sqrt{314.35}\hfill \\ x\approx 17.7\text{miles}\hfill \end{array}[/latex]

The boat is about 17.7 miles from port.