The three angles must add up to 180 degrees. From this, we can determine that

[latex]\begin{align} \beta =180^\circ -50^\circ -30^\circ =100^\circ \end{align}[/latex]

To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle [latex]\alpha =50^\circ[/latex] and its corresponding side [latex]a=10[/latex]. We can use the following proportion from the Law of Sines to find the length of [latex]c[/latex].

[latex]\begin{align}&\frac{\sin \left(50^\circ \right)}{10}=\frac{\sin \left(30^\circ \right)}{c} \\ &c\frac{\sin \left(50^\circ \right)}{10}=\sin \left(30^\circ \right) && \text{Multiply both sides by }c. \\ &c=\sin \left(30^\circ \right)\frac{10}{\sin \left(50^\circ \right)} && \text{Multiply by the reciprocal to isolate }c. \\ &c\approx 6.5 \end{align}[/latex]

Similarly, to solve for [latex]b[/latex], we set up another proportion.

[latex]\begin{align} &\frac{\sin \left(50^\circ \right)}{10}=\frac{\sin \left(100^\circ \right)}{b} \\ &b\sin \left(50^\circ \right)=10\sin \left(100^\circ \right) && \text{Multiply both sides by }b. \\ &b=\frac{10\sin \left(100^\circ \right)}{\sin \left(50^\circ \right)} && \text{Multiply by the reciprocal to isolate }b. \\ &b\approx 12.9\end{align}[/latex]

Therefore, the complete set of angles and sides is

[latex]\begin{gathered} \alpha =50^\circ,\beta =100^\circ,\gamma =30^\circ \\ a=10,b\approx 12.9,c\approx 6.5 \end{gathered}[/latex]

Use the Law of Sines to find angle [latex]\beta[/latex] and angle [latex]\gamma[/latex], and then side [latex]c[/latex]. Solving for [latex]\beta[/latex], we have the proportion

[latex]\begin{gathered} \frac{\sin \alpha }{a}=\frac{\sin \beta }{b} \\ \frac{\sin \left(35^\circ \right)}{6}=\frac{\sin \beta }{8}\\ \frac{8\sin \left(35^\circ \right)}{6}=\sin \beta \\ 0.7648\approx \sin \beta \\ {\sin }^{-1}\left(0.7648\right)\approx 49.9^\circ \\ \beta \approx 49.9^\circ \end{gathered}[/latex]

However, in the diagram, angle [latex]\beta[/latex] appears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement of [latex]\beta ?[/latex] Let’s investigate further. Dropping a perpendicular from [latex]\gamma[/latex] and viewing the triangle from a right angle perspective, we have Figure 11. It appears that there may be a second triangle that will fit the given criteria.

Figure 11

The angle supplementary to [latex]\beta[/latex] is approximately equal to 49.9°, which means that [latex]\beta =180^\circ -49.9^\circ =130.1^\circ[/latex]. (Remember that the sine function is positive in both the first and second quadrants.) Solving for [latex]\gamma[/latex], we have

[latex]\gamma =180^\circ -35^\circ -130.1^\circ \approx 14.9^\circ[/latex]

We can then use these measurements to solve the other triangle. Since [latex]{\gamma }^{\prime }[/latex] is supplementary to [latex]\gamma[/latex], we have

[latex]{\gamma }^{\prime }=180^\circ -35^\circ -49.9^\circ \approx 95.1^\circ[/latex]

Now we need to find [latex]c[/latex] and [latex]{c}^{\prime }[/latex].

We have

[latex]\begin{gathered}\frac{c}{\sin \left(14.9^\circ \right)}=\frac{6}{\sin \left(35^\circ \right)} \\ c=\frac{6\sin \left(14.9^\circ \right)}{\sin \left(35^\circ \right)}\approx 2.7 \end{gathered}[/latex]

Finally,

[latex]\begin{gathered}\frac{{c}^{\prime }}{\sin \left(95.1^\circ \right)}=\frac{6}{\sin \left(35^\circ \right)} \\ {c}^{\prime }=\frac{6\sin \left(95.1^\circ \right)}{\sin \left(35^\circ \right)}\approx 10.4 \end{gathered}[/latex]

To summarize, there are two triangles with an angle of 35°, an adjacent side of 8, and an opposite side of 6, as shown in Figure 12.

Figure 12

However, we were looking for the values for the triangle with an obtuse angle [latex]\beta[/latex]. We can see them in the first triangle (a) in Figure 12.

In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle [latex]\gamma =85^\circ[/latex], and its corresponding side [latex]c=12[/latex], and we know side [latex]b=9[/latex]. We will use this proportion to solve for [latex]\beta[/latex].

[latex]\begin{align}\frac{\sin \left(85^\circ \right)}{12}&=\frac{\sin \beta }{9} && \text{Isolate the unknown}.\\ \frac{9\sin \left(85^\circ \right)}{12}&=\sin \beta \end{align}[/latex]

To find [latex]\beta[/latex], apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for [latex]\beta[/latex]. It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions.

[latex]\begin{align}\beta &={\sin }^{-1}\left(\frac{9\sin \left(85^\circ \right)}{12}\right) \\ \beta &\approx {\sin }^{-1}\left(0.7471\right) \\ \beta &\approx 48.3^\circ \end{align}[/latex]

In this case, if we subtract [latex]\beta[/latex] from 180°, we find that there may be a second possible solution. Thus, [latex]\beta =180^\circ -48.3^\circ \approx 131.7^\circ[/latex]. To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives

[latex]\alpha =180^\circ -85^\circ -131.7^\circ \approx -36.7^\circ[/latex],

which is impossible, and so [latex]\beta \approx 48.3^\circ[/latex].

To find the remaining missing values, we calculate [latex]\alpha =180^\circ -85^\circ -48.3^\circ \approx 46.7^\circ[/latex]. Now, only side [latex]a[/latex] is needed. Use the Law of Sines to solve for [latex]a[/latex] by one of the proportions.

[latex]\begin{gathered} \frac{\sin \left(85^\circ \right)}{12}=\frac{\sin \left(46.7^\circ \right)}{a} \\ a\frac{\sin \left(85^\circ \right)}{12}=\sin \left(46.7^\circ \right) \\ a=\frac{12\sin \left(46.7^\circ \right)}{\sin \left(85^\circ \right)}\approx 8.8 \end{gathered}[/latex]

The complete set of solutions for the given triangle is

[latex]\begin{gathered} \alpha \approx 46.7^\circ \text{, }a\approx 8.8 \\ \beta \approx 48.3^\circ \text{, }b=9 \\ \gamma =85^\circ \text{, }c=12\end{gathered}[/latex]

Using the given information, we can solve for the angle opposite the side of length 10.

[latex]\begin{gathered}\frac{\sin \alpha }{10}=\frac{\sin \left(50^\circ \right)}{4} \\ \sin \alpha =\frac{10\sin \left(50^\circ \right)}{4} \\ \sin \alpha \approx 1.915 \end{gathered}[/latex]

Figure 14

We can stop here without finding the value of [latex]\alpha[/latex]. Because the range of the sine function is [latex]\left[-1,1\right][/latex], it is impossible for the sine value to be 1.915. In fact, inputting [latex]{\sin }^{-1}\left(1.915\right)[/latex] in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions.

Using the formula, we have

[latex]\begin{align}\text{Area}&=\frac{1}{2}ab\sin \gamma \\ \text{Area}&=\frac{1}{2}\left(90\right)\left(52\right)\sin \left(102^\circ \right) \\ \text{Area}&\approx 2289\text{square units} \end{align}[/latex]

To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side [latex]a[/latex], and then use right triangle relationships to find the height of the aircraft, [latex]h[/latex].

Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180°−15°−35°=130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship.

[latex]\begin{gathered} \frac{\sin \left(130^\circ \right)}{20}=\frac{\sin \left(35^\circ \right)}{a} \\ a\sin \left(130^\circ \right)=20\sin \left(35^\circ \right) \\ a=\frac{20\sin \left(35^\circ \right)}{\sin \left(130^\circ \right)} \\ a\approx 14.98 \end{gathered}[/latex]

The distance from one station to the aircraft is about 14.98 miles.

Now that we know [latex]a[/latex], we can use right triangle relationships to solve for [latex]h[/latex].

[latex]\begin{gathered}\sin \left(15^\circ \right)=\frac{\text{opposite}}{\text{hypotenuse}} \\ \sin \left(15^\circ \right)=\frac{h}{a} \\ \sin \left(15^\circ \right)=\frac{h}{14.98} \\ h=14.98\sin \left(15^\circ \right) \\ h\approx 3.88 \end{gathered}[/latex]

The aircraft is at an altitude of approximately 3.9 miles.

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 22

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 23

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 24

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.