First, make note of what is given: two sides and the angle between them. This arrangement is classified as SAS and supplies the data needed to apply the Law of Cosines.

Each one of the three laws of cosines begins with the square of an unknown side opposite a known angle. For this example, the first side to solve for is side [latex]b[/latex], as we know the measurement of the opposite angle [latex]\beta[/latex].

[latex]\begin{align}&{b}^{2}={a}^{2}+{c}^{2}-2ac\cos \beta \\ &{b}^{2}={10}^{2}+{12}^{2}-2\left(10\right)\left(12\right)\cos \left({30}^{\circ }\right) && \text{Substitute the measurements for the known quantities}. \\ &{b}^{2}=100+144 - 240\left(\frac{\sqrt{3}}{2}\right)&& \text{Evaluate the cosine and begin to simplify}. \\ &{b}^{2}=244 - 120\sqrt{3} \\ &b=\sqrt{244 - 120\sqrt{3}}&& \text{Use the square root property}. \\ &b\approx 6.013 \end{align}[/latex]

Because we are solving for a length, we use only the positive square root. Now that we know the length [latex]b[/latex], we can use the Law of Sines to fill in the remaining angles of the triangle. Solving for angle [latex]\alpha[/latex], we have

[latex]\begin{align}&\frac{\sin \alpha }{a}=\frac{\sin \beta }{b} \\ &\frac{\sin \alpha }{10}=\frac{\sin \left(30^\circ \right)}{6.013} \\ &\sin \alpha =\frac{10\sin \left(30^\circ \right)}{6.013}&& \text{Multiply both sides of the equation by 10}. \\ &\alpha ={\sin }^{-1}\left(\frac{10\sin \left(30^\circ \right)}{6.013}\right)&& \text{Find the inverse sine of }\frac{10\sin \left(30^\circ \right)}{6.013}. \\ &\alpha \approx 56.3^\circ \end{align}[/latex]

The other possibility for [latex]\alpha[/latex] would be [latex]\alpha =180^\circ -56.3^\circ \approx 123.7^\circ[/latex]. In the original diagram, [latex]\alpha[/latex] is adjacent to the longest side, so [latex]\alpha[/latex] is an acute angle and, therefore, [latex]123.7^\circ[/latex] does not make sense. Notice that if we choose to apply the Law of Cosines, we arrive at a unique answer. We do not have to consider the other possibilities, as cosine is unique for angles between [latex]0^\circ[/latex] and [latex]180^\circ[/latex]. Proceeding with [latex]\alpha \approx 56.3^\circ[/latex], we can then find the third angle of the triangle.

[latex]\gamma =180^\circ -30^\circ -56.3^\circ \approx 93.7^\circ[/latex]

The complete set of angles and sides is

[latex]\begin{align}&\alpha \approx 56.3^\circ && a=10\\ &\beta =30^\circ && b\approx 6.013\\ &\gamma \approx 93.7^\circ && c=12 \end{align}[/latex]

For this example, we have no angles. We can solve for any angle using the Law of Cosines. To solve for angle [latex]\alpha[/latex], we have

[latex]\begin{align} &{a}^{2}={b}^{2}+{c}^{2}-2bc\cos \alpha \\ &{20}^{2}={25}^{2}+{18}^{2}-2\left(25\right)\left(18\right)\cos \alpha&& \text{Substitute the appropriate measurements}. \\ &400=625+324 - 900\cos \alpha&& \text{Simplify in each step}. \\ &400=949 - 900\cos \alpha \\ &-549=-900\cos \alpha&& \text{Isolate cos }\alpha . \\ &\frac{-549}{-900}=\cos \alpha \\ &0.61\approx \cos \alpha \\ &{\cos }^{-1}\left(0.61\right)\approx \alpha&& \text{Find the inverse cosine}. \\ &\alpha \approx 52.4^\circ \end{align}[/latex]

See Figure 5.

Figure 5

Analysis of the Solution

Because the inverse cosine can return any angle between 0 and 180 degrees, there will not be any ambiguous cases using this method.

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

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{gathered}{a}^{2}={b}^{2}+{c}^{2}-2bc\cos \theta \\ {\left(2420\right)}^{2}={\left(5050\right)}^{2}+{\left(6000\right)}^{2}-2\left(5050\right)\left(6000\right)\cos \theta \\ {\left(2420\right)}^{2}-{\left(5050\right)}^{2}-{\left(6000\right)}^{2}=-2\left(5050\right)\left(6000\right)\cos \theta \\ \frac{{\left(2420\right)}^{2}-{\left(5050\right)}^{2}-{\left(6000\right)}^{2}}{-2\left(5050\right)\left(6000\right)}=\cos \theta \\ \cos \theta \approx 0.9183 \\ \theta \approx {\cos }^{-1}\left(0.9183\right) \\ \theta \approx 23.3^\circ \end{gathered}[/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.

Figure 7

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

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

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

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

The boat is about 17.7 miles from port.

First, we calculate [latex]s[/latex].

[latex]\begin{align} s&=\frac{\left(a+b+c\right)}{2} \\ s&=\frac{\left(10+15+7\right)}{2}=16 \end{align}[/latex]

Then we apply the formula.

[latex]\begin{align} \text{Area}&=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ \text{Area}&=\sqrt{16\left(16 - 10\right)\left(16 - 15\right)\left(16 - 7\right)} \\ \text{Area}&\approx 29.4 \end{align}[/latex]

The area is approximately 29.4 square units.

Find the measurement for [latex]s[/latex], which is one-half of the perimeter.

[latex]\begin{align}s&=\frac{\left(62.4+43.5+34.1\right)}{2} \\ s&=70\text{m} \end{align}[/latex]

Apply Heron’s formula.

[latex]\begin{align}\text{Area}&=\sqrt{70\left(70 - 62.4\right)\left(70 - 43.5\right)\left(70 - 34.1\right)} \\ \text{Area}&=\sqrt{506,118.2} \\ \text{Area}&\approx 711.4 \end{align}[/latex]

The developer has about 711.4 square meters.