This is a very thorough section of all cases for the Law of Sines and their applications. Please pay close attention to my tabular method. While it is terribly algorithmic, students tend to do much better with it than other methods I have tried.
The following is a list of learning objectives for this section.
Learning Objectives (click to expand)
Use the Law of Sines to find a missing side in an oblique triangle.
Solve an application using the Law of Sines.
Use the Law of Sines to find all possible solutions for the ambiguous case.
Solve an application involving the ambiguous case.
We have learned to use trigonometric ratios to solve right triangles. However, the ratios are only valid for the sides of right triangles. Can we find unknown sides or angles in an oblique triangle?
Figure \(\PageIndex{1}\)
In this section and the next, we find relationships between the sides and angles of oblique triangles. These relationships are called the Law of Sines and the Law of Cosines. We use what we already know about right triangles to derive these new rules.
The Law of Sines
Theorem: Law of Sines
If the angles of a triangle are \(A\), \(B\), and \(C\), and the opposite sides are \(a\), \(b\), and \(c\), respectively, then
\[\dfrac{\sin\left( A \right)}{a}=\dfrac{\sin\left( B \right)}{b}=\dfrac{\sin\left( C \right)}{c},\nonumber \]or equivalently,\[\dfrac{a}{\sin\left( A \right)}=\dfrac{b}{\sin\left( B \right)}=\dfrac{c}{\sin\left( C \right)}.\nonumber \]
Proof
Consider the oblique triangle in the figure below.
By drawing in the altitude \(h\) of the triangle, we create two right triangles, \(\triangle B C D\) and \(\triangle A B D\). We can write expressions in terms of \(h\) for \(\sin \left(A\right)\) and for \(\sin \left(C\right)\).
Looking at \(\triangle B C D\), we see that\[\dfrac{h}{a}=\sin\left( C\right).\nonumber \]Looking at \(\triangle A B D\), we see that\[\dfrac{h}{c}=\sin \left(A\right).\nonumber \]Now we solve each of these equations for \(h\):\[ \begin{array}{crcl}
\quad & \dfrac{h}{a} = \sin\left( C \right) & \implies & h = a \, \sin\left( C \right) \\[6pt] \text{and} & \dfrac{h}{c}=\sin\left( A \right) & \implies & h = c \, \sin\left( A \right) \\[6pt] \end{array} \nonumber \]Therefore,\[ \begin{array}{rrclcl}
& a \, \sin\left( C \right) & = & c \, \sin\left( A \right) & & \\[6pt] \implies & \dfrac{\sin\left( C \right)}{c} & = & \dfrac{\sin\left( A \right)}{a} & \quad & \left( \text{divide both sides by }ac \right) \\[6pt] \end{array} \nonumber \]We have derived a relationship between the angles \(A\) and \(C\) and their opposite sides, \(a\) and \(c\). If we know any three of these quantities, we can find the fourth.
In a similar way, by drawing in the altitude from the vertex \(C\), we can show that\[\dfrac{\sin\left( A \right)}{a}=\dfrac{\sin\left( B \right)}{b}. \nonumber \]Putting both results together, we have proved the Law of Sines.
The Law of Sines is true for any triangle, whether acute, right, or obtuse. In the following example, we use the Law of Sines to find a distance.
Example \(\PageIndex{1}\)
Two observers on the shore sight a ship at an unknown distance from the shoreline. The observers are 400 yards apart at points \(A\) and \(B\), and they each measure the angle from the shoreline to the ship, as shown below. How far is the ship from the observer at \(A\)?
Figure \(\PageIndex{2}\)
Solution
First, note that \(\triangle ABC\) is not a right triangle, so we cannot use the trigonometric ratios directly to find the sides of this triangle.
Let \(d\) be the unknown distance opposite \(\angle B=79.4^{\circ}\). To use the Law of Sines, we must know another angle and the side opposite that angle. Luckily, we know that side \(c = 400\), and we can compute the angle at the ship, \(\angle C\).\[\angle C=180^{\circ}-\left(79.4^{\circ}+83.2^{\circ}\right)=17.4^{\circ}.\nonumber\]Now we apply the Law of Sines, using angles \(B\) and \(C\).\[\begin{array}{rrclcl} & \dfrac{b}{\sin\left(B\right)} & = & \dfrac{c}{\sin\left(C\right)} & \quad & \left(\text{Law of Sines}\right) \\[6pt] \implies & \dfrac{d}{\sin\left(79.4^{\circ}\right)} & = & \dfrac{400}{\sin\left(17.4^{\circ}\right)} & \quad & \left(\text{substitute}\right) \\[6pt] \implies & d & = & \dfrac{400\sin\left(79.4^{\circ}\right)}{\sin\left(17.4^{\circ}\right)} & \quad & \left(\text{multiply both sides by }\sin\left(79.4^{\circ}\right)\right) \\[6pt] \implies & d & \approx & 1315 & & \\[6pt] \end{array}\nonumber\]The ship is about 1315 yards from the observer at \(A\).
In Example \(\PageIndex{1}\), we saved all calculator work for the last step. That is, we did not approximate the values of \(\sin\left(79.4^{\circ}\right)\) and \(\sin\left(17.4^{\circ}\right)\), and then use those approximations to get a further approximation—doing so would propagate errors.
Advice: Save Calculator Work for the Last Step
As much as possible, only reach for technology in any mathematical work at the very last moment.
Another piece of advice is to collect information into tables when working through an application. This habit is especially helpful with the current material (and the material in the next section).
In Example \(\PageIndex{1}\), we could have collected the information into a table where the columns are the angles and their opposite side lengths. When missing information, we place a temporary question mark. Therefore, we would have started with the table\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline \angle A = 83.2^{\circ} & a = ? \\[6pt] \hline \angle B = 79.4^{\circ} & b = ? \\[6pt] \hline \angle C = ? & c = 400 \\[6pt] \hline \end{array}\nonumber\]Looking at that first column, we realize these are the interior angles of a triangle and, as such, should sum to \(180^{\circ}\). Hence,\[\angle C = 180^{\circ} - 83.2^{\circ} - 79.4^{\circ} = 17.4^{\circ}.\nonumber\]Thus, our table becomes\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline \angle A = 83.2^{\circ} & a = ? \\[6pt] \hline \angle B = 79.4^{\circ} & b = ? \\[6pt] \hline \angle C = 17.4^{\circ} & c = 400 \checkmark \\[6pt] \hline \end{array}\nonumber\]That checkmark (\(\checkmark\)) is not an accident. Recall, the Law of Sines tells us\[\dfrac{a}{\sin\left(A\right)}=\dfrac{b}{\sin\left(B\right)}=\dfrac{c}{\sin\left(C\right)}.\nonumber\]If we have the sine of an angle and the length of its opposite side, then we know the value of these ratios.
That is, once we have a complete row in our table, we can try to use the Law of Sines (I say "try" because sometimes solving a triangle using the Law of Sines will not work. We will see those cases momentarily). Since our goal is to find the side length opposite \(\angle B\) (which, for the sake of building the table, I call \(b\)), we work with the equation\[\dfrac{b}{\sin\left(B\right)}=\dfrac{c}{\sin\left(C\right)}.\nonumber\]We substitute the values from the table to get\[\dfrac{b}{\sin\left(79.4^{\circ}\right)} = \dfrac{400}{\sin\left(17.4^{\circ}\right)}\nonumber\]and finish as we did in Example \(\PageIndex{1}\).
Checkpoint \(\PageIndex{1}\)
Delbert and Francine are 40 feet apart on one side of a river. They make angle measurements to a pine tree on the opposite shore, as shown below. What is the distance from Francine to the pine tree?
Figure \(\PageIndex{3}\)
Answer
About 114.8 feet
Solving Triangles with the Law of Sines
To apply the Law of Sines to find a side, we must know one angle of the triangle and its opposite side (either \(a\) and \(A\), or \(b\) and \(B\), or \(c\) and \(C\)), and one other angle. Then, we can find the side opposite that angle.
Example \(\PageIndex{2}\)
In the triangle shown below, \(A=37^{\circ}\), \(B=54^{\circ}\), and \(a=11\).
Find \(b\).
Solve the triangle.
Figure \(\PageIndex{4}\)
Solutions
We begin by building our table of given information (using question marks where we aren't explicitly given information).\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline A = 37^{\circ} & a = 11 \checkmark \\[6pt] \hline B = 54^{\circ} & b = ? \\[6pt] \hline C = ? & c = ? \\[6pt] \hline \end{array}\nonumber\]We already have a complete row! This means we can immediately try using the Law of Sines; however, I cannot resist finding the value of \(C\) when it is so easy to compute.\[C = 180^{\circ} - 54^{\circ} - 37^{\circ} = 89^{\circ}.\nonumber\]Hence, our table becomes\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline A = 37^{\circ} & a = 11 \checkmark \\[6pt] \hline B = 54^{\circ} & b = ? \\[6pt] \hline C = 89^{\circ} & c = ? \\[6pt] \hline \end{array}\nonumber\]
We use the Law of Sines with \(a\) and angle \(A\) to find \(b\).\[\begin{array}{rrclcl} & \dfrac{a}{\sin\left(A\right)} & = & \dfrac{b}{\sin\left(B\right)} & \quad & \left(\text{Law of Sines}\right) \\[6pt] \implies & \dfrac{11}{\sin\left(37^{\circ}\right)} & = & \dfrac{b}{\sin\left(54^{\circ}\right)} & \quad & \left(\text{substitute}\right) \\[6pt] \implies & \dfrac{11\sin\left(54^{\circ}\right)}{\sin\left(37^{\circ}\right)} & = & b & \quad & \left(\text{multiply both sides by }\sin\left(54^{\circ}\right)\right) \\[6pt] \end{array}\nonumber\]Evaluating this expression with a calculator, we find that \(b \approx 14.79\). We fill this in our table for display purposes only (we will not use the rounded value for computing).\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline A = 37^{\circ} & a = 11 \checkmark \\[6pt] \hline B = 54^{\circ} & b \approx 14.79 \\[6pt] \hline C = 89^{\circ} & c = ? \\[6pt] \hline \end{array}\nonumber\]
At this point, we are very close to having the triangle solved. We only need the length of side \(c\). Using the given side, \(a\), is safer than using the value we calculated for \(b\).\[\begin{array}{rrclcl} & \dfrac{a}{\sin\left(A\right)} & = & \dfrac{c}{\sin\left(C\right)} & \quad & \left(\text{Law of Sines}\right) \\[6pt] \implies & \dfrac{11}{\sin\left(37^{\circ}\right)} & = & \dfrac{c}{\sin\left(89^{\circ}\right)} & \quad & \left(\text{substitute}\right) \\[6pt] \implies & \dfrac{11\sin\left(89^{\circ}\right)}{\sin\left(37^{\circ}\right)} & = & c & \quad & \left(\text{multiply both sides by }\sin\left(89^{\circ}\right)\right) \\[6pt] \end{array}\nonumber\]So \(c \approx 18.28\). Now our table is complete, and we have solved the triangle.\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline A = 37^{\circ} & a = 11 \checkmark \\[6pt] \hline B = 54^{\circ} & b \approx 14.79 \\[6pt] \hline C = 89^{\circ} & c \approx 18.28 \\[6pt] \hline \end{array}\nonumber\]
Checkpoint \(\PageIndex{2}\)
In the triangle below, \(A=65^{\circ}\), \(C=42^{\circ}\), and \(c=16\). Solve the triangle.
Figure \(\PageIndex{5}\)
Answer
\(B=73^{\circ}, b \approx 22.87, a \approx 21.68\)
Finding an Angle
We can also use the Law of Sines to find an unknown angle of a triangle. We must know two sides of the triangle and the angle opposite one of them.
Example \(\PageIndex{3}\)
In the triangle below, \(B=55^{\circ}\), \(a=5\), and \(b=11\). Solve the triangle.
Figure \(\PageIndex{6}\)
Solution
Again, let's start by filling out a table.\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline A = ? & a = 5 \\[6pt] \hline B = 55^{\circ} & b = 11 \checkmark \\[6pt] \hline C = ? & c = ? \\[6pt] \hline \end{array}\nonumber\]This is the first time where we cannot simply find the remaining angles using the fact that the sum of angles in a triangle is \(180^{\circ}\). Luckily, we still have a complete row so we can use the Law of Sines, and we will do so to find \(\sin\left(A\right)\).\[\begin{array}{rrclcl} & \dfrac{\sin\left(A\right)}{a} & = & \dfrac{\sin\left(B\right)}{b} & \quad & \left(\text{Law of Sines}\right) \\[6pt] \implies & \dfrac{\sin\left(A\right)}{5} & = & \dfrac{\sin\left(55^{\circ}\right)}{11} & \quad & \left(\text{substitute}\right) \\[6pt] \implies & \sin\left(A\right) & = & \dfrac{5\sin\left(55^{\circ}\right)}{11} & \quad & \left(\text{multiply both sides by }5\right) \\[6pt] \implies & \widehat{A} & = & \sin^{-1}\left(\dfrac{5\sin\left(55^{\circ}\right)}{11}\right) & \quad & \left(\text{finding the reference angle - }\right. \\[6pt] & & & & & \left.\text{see }\textbf{Unpacking Law of Sines Subtleties}\right) \\[6pt] \end{array}\nonumber\]We have a reference angle, and we know that each angle in a triangle must be positive but less than \(180^{\circ}\). Therefore, \(A\) is either in \(\mathrm{QI}\) or \(\mathrm{QII}\). In all honesty, since \(A\) is an angle within a triangle and not an angle in standard position within a prescribed coordinate system, it is incorrect to say that \(A\in\mathrm{QI}\) or \(A\in\mathrm{QII}\); however, I am okay with the "looseness" of this language.
Here's where a little extra logic comes into play. Since the length of side \(b\) is greater than the length of side \(a\), it must be the case that \(B \gt A\). Since \(B = 55^{\circ}\), \(0 \lt A \lt 55^{\circ}\). So\[A = \sin^{-1}\left(\dfrac{5\sin\left(55^{\circ}\right)}{11}\right) \approx 21.9^{\circ}.\nonumber\]Let's update our table.\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline A \approx 21.9^{\circ} & a = 5 \\[6pt] \hline B = 55^{\circ} & b = 11 \checkmark \\[6pt] \hline C = ? & c = ? \\[6pt] \hline \end{array}\nonumber\]Now that we know two angles, we can find the third.\[C = 180^{\circ} - B - A \approx 103.1^{\circ}.\nonumber\]Again, we update the table.\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline A \approx 21.9^{\circ} & a = 5 \\[6pt] \hline B = 55^{\circ} & b = 11 \checkmark \\[6pt] \hline C \approx 103.1^{\circ} & c = ? \\[6pt] \hline \end{array}\nonumber\]Finally, we use the Law of Sines to find side \(c\).\[\begin{array}{rrclcl} & \dfrac{b}{\sin\left(B\right)} & = & \dfrac{c}{\sin\left(C\right)} & \quad & \left(\text{Law of Sines}\right) \\[6pt] \implies & \dfrac{b\,\sin\left(C\right)}{\sin\left(B\right)} & = & c & \quad & \left(\text{multiplying both sides by }\sin\left(C\right)\right) \\[6pt] \end{array}\nonumber\]Substituting in the values of \(b\), \(B\), and \(C\) (stored in the calculator) and evaluating this expression with a calculator gives \(c \approx 13.1\). Thus, our final table is the following:\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline A \approx 21.9^{\circ} & a = 5 \\[6pt] \hline B = 55^{\circ} & b = 11 \checkmark \\[6pt] \hline C \approx 103.1^{\circ} & c \approx 13.1 \\[6pt] \hline \end{array}\nonumber\]
Caution: Do Not Use Rounded Values to Calculate
Example \(\PageIndex{3}\) has a lot going on; however, the simplest thing to cover at this point is proper etiquette when it comes to calculations.
While I am displaying the approximations in the tables within Example \(\PageIndex{3}\), any computations relying on these are done using the exact or stored values in the calculator. For example, when I do any computation relying upon \(A\), I either use \(\sin^{-1}\left(\frac{5\sin\left(55^{\circ}\right)}{11}\right)\) or the unrounded approximation stored in my calculator. This is done to avoid the propagation of rounding errors. All modern calculators can store results, and I suggest you learn how to use that feature. For example, on the calculator I use (the TI-30XS), there is a button labeled \(\fbox{$\mathrm{sto}\rightarrow$}\) and another button labeled \(\fbox{$\mathrm{X}^{\mathrm{yzt}}_{\mathrm{abc}}$}\). To store the result of \(\sin^{-1}\left(\frac{5\sin\left(55^{\circ}\right)}{11}\right)\), I use the following button sequence:\[\fbox{$\mathrm{sto}\rightarrow$} \quad \fbox{$\mathrm{X}^{\mathrm{yzt}}_{\mathrm{abc}}$}\,\left(\text{tapping this a few times until }a\text{ is displayed}\right) \quad \fbox{$\mathrm{enter}$}\nonumber\]When I need to access that stored value, I tap \(\fbox{$\mathrm{X}^{\mathrm{yzt}}_{\mathrm{abc}}$}\) until I get the variable I stored it as, and then hit \(\fbox{$\mathrm{enter}$}\).
Checkpoint \(\PageIndex{3}\)
Sketch a triangle with \(C=93^{\circ}\), \(a=7\), and \(c=11\).
Use the Law of Sines to find another angle of the triangle.
Solve the triangle, and label your sketch with the results.
Answers
\(A \approx 39.5^{\circ}\)
\(b \approx 8.13, B \approx 47.5^{\circ}\)
In Examples \(\PageIndex{1}\) and \(\PageIndex{2}\), we used the Law of Sines to find missing side lengths, but we didn't need the Law of Sines to find angles. In Example \(\PageIndex{3}\) (and Checkpoint \(\PageIndex{3}\)), on the other hand, we had no choice but to use the Law of Sines to find a missing angle—this is where we need to be careful when using the Law of Sines.
Note: Unpacking Law of Sines Subtleties
It's imperative to recognize that all angles within a triangle must be positive but less than \(180^{\circ}\). Thus, the sine of any angle within a triangle will always be positive. Since the side lengths are also positive, the arcsine we use to find the missing angle will always return an acute angle (i.e., an angle between \(0^{\circ}\) and \(90^{\circ}\)).
"Why do we need to be aware of this?"
When using the Law of Sines to find a missing angle in a triangle, there could be two such triangles satisfying the given conditions—one containing the acute returned by the arcsine, and the other containing its supplement.
"How do we know when we have two triangles for a given set of information?"
When using the Law of Sines to find a missing angle, we consider the angle returned by the arcsine to be a reference angle. To help with this discussion, let's call this reference angle \(\widehat{\theta}\).
\(\widehat{\theta}\) gives two possibilities for the true value of the angle we seek. The first is \(\theta = \widehat{\theta}\). The second is its supplement, \(\theta^{\prime} = 180^{\circ} - \widehat{\theta}\) (the "prime" notation is traditionally used for the supplementary angle). While \(\theta\) is the solution that gives one triangle, \(\theta^{\prime}\) is a candidate for a second value (and, therefore, a second triangle). This is because the sine is positive and, therefore, its argument could also be between \(90^{\circ}\) and \(180^{\circ}\).
With our candidate angles, \(\theta\) and \(\theta^{\prime}\), in hand, we then check whether the other given angle works with each of these computed values. Sometimes, the provided information works with both candidate values; however, at the very least, the acute angle always works.
To illustrate this concept, let's look back at Example \(\PageIndex{3}\). We found \(\widehat{A} = \sin^{-1}\left(\frac{5\sin\left(55^{\circ}\right)}{11}\right) \approx 21.9^{\circ}\). Thus, we immediately knew \(A \approx 21.9^{\circ}\) worked and would yield a triangle; however, what about the supplement of this angle?
In \(\mathrm{QII}\), our alternate option for \(A\) is \(A^{\prime} = 180^{\circ} - \widehat{A} \approx 158.1^{\circ}\); however, if we tried to fit this obtuse angle into a triangle, we would compute the final missing angle to be\[C = 180^{\circ} - A - B \approx 180^{\circ} - 158.1^{\circ} - 55^{\circ} = -33.1^{\circ}.\nonumber\]This is impossible because angles within a triangle are never negative. Thus, we only have one triangle (given by the acute angle \(A \approx 21.9^{\circ}\)).
Again, sometimes both angles produce (different) triangles, and sometimes only the acute angle will work. You should always check whether both angles provide solutions.
Because there can be more than one solution for a triangle when using the Law of Sines to find a missing angle, we call this situation the ambiguous case for the Law of Sines.
Example \(\PageIndex{4}\)
Solve the triangle in which \(B=14.4^{\circ}\), \(a=8\), and \(b=3\). Sketch the resulting triangle.
Solution
We start, as usual, by building a table.\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline A = ? & a = 8 \\[6pt] \hline B = 14.4^{\circ} & b = 3 \checkmark \\[6pt] \hline C = ? & c = ? \\[6pt] \hline \end{array}\nonumber\]Other than the numbers, this looks like it's going to mimic what we did in Example \(\PageIndex{3}\).\[\begin{array}{rrclcl} & \dfrac{\sin\left(A\right)}{a} & = & \dfrac{\sin\left(B\right)}{b} & \quad & \left(\text{Law of Sines}\right) \\[6pt] \implies & \sin\left(A\right) & = & \dfrac{a\,\sin\left(B\right)}{b} & \quad & \left(\text{multiplying both sides by }a\right) \\[6pt] \implies & \widehat{A} & = & \sin^{-1}\left(\dfrac{a\sin\left(B\right)}{b}\right) & \quad & \left(\text{finding the reference angle}\right) \\[6pt] \end{array}\nonumber\]Grabbing a calculator, substituting in the known values of \(B\), \(a\), and \(b\), we get \(\widehat{A} \approx 41.5^{\circ}\) (again, the exact value is stored in the calculator to be used for other calculations). As stated previously, the acute angle will always work because if \(A \approx 41.5^{\circ}\), then \(C = 180^{\circ} - A - B \approx 124.1^{\circ}\); however, before throwing all of this into a table, let's check to see if the supplementary (obtuse) angle works as well. Let\[A^{\prime} = 180^{\circ} - A \approx 138.5^{\circ}.\nonumber\]Then\[C^{\prime} = 180^{\circ} - A^{\prime} - B \approx 27.1^{\circ}.\nonumber\]Since this is not a negative angle, this creates another workable triangle!
Thus, if \(A \approx 41.5^{\circ}\), there is room in the triangle for the final, third angle \(C \approx 124.1^{\circ}\), and if \(A^{\prime} \approx 138.5^{\circ}\), there is room in another triangle for the final, third angle \(C^{\prime} \approx 27.1^{\circ}\). In these ambiguous cases, we build two tables of information, one for the acute angle and one for the obtuse angle.\[\begin{array}{ccc} \begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline A \approx 41.5^{\circ} & a = 8 \\[6pt] \hline B = 14.4^{\circ} & b = 3 \checkmark \\[6pt] \hline C \approx 124.1^{\circ} & c = ? \\[6pt] \hline \end{array} & \quad & \begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline A^{\prime} \approx 138.5^{\circ} & a = 8 \\[6pt] \hline B = 14.4^{\circ} & b = 3 \checkmark \\[6pt] \hline C^{\prime} \approx 27.1^{\circ} & c^{\prime} = ? \\[6pt] \hline \end{array} \\[6pt] \end{array}\nonumber\]In either case, we need to compute the final missing sides, \(c\) and \(c^{\prime}\), using the Law of Sines. Remember, when using the Law of Sines, we want to use as much of the given information as possible because it's cleaner than our approximated information. Moreover, we use the non-rounded values and save rounding for the end.\[\begin{array}{rrclcl} & \dfrac{c}{\sin\left(C\right)} & = & \dfrac{b}{\sin\left(B\right)} & \quad & \left(\text{Law of Sines}\right) \\[6pt] \implies & c & = & \dfrac{b\,\sin\left(C\right)}{\sin\left(B\right)} & \quad & \left(\text{multiply both sides by }\sin\left(C\right)\right) \\[6pt] \end{array}\nonumber\]This formula is the same for \(c^{\prime}\) (just replace \(C\) with \(C^{\prime}\)). Hence, we grab a calculator, substitute the necessary values, and get our approximations (listed in the tables below).\[\begin{array}{ccc} \begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline A \approx 41.5^{\circ} & a = 8 \\[6pt] \hline B = 14.4^{\circ} & b = 3 \checkmark \\[6pt] \hline C \approx 124.1^{\circ} & c \approx 9.99 \\[6pt] \hline \end{array} & \quad & \begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline A^{\prime} \approx 138.5^{\circ} & a = 8 \\[6pt] \hline B = 14.4^{\circ} & b = 3 \checkmark \\[6pt] \hline C^{\prime} \approx 27.1^{\circ} & c^{\prime} \approx 5.5 \\[6pt] \hline \end{array} \\[6pt] \end{array}\nonumber\]The following figure shows both triangles.
Figure \(\PageIndex{7}\)
Checkpoint \(\PageIndex{4}\)
Suppose that \(C=29.7^{\circ}\), \(b=8\), and \(c=5\).
Find two possible values for angle \(B\).
Solve the triangle for both values of \(B\), and sketch both solutions.
Answers
\(B=52.4^{\circ}\) or \(B=127.6^{\circ}\)
\(A=97.9^{\circ}, a=10\)
Figure \(\PageIndex{8}\)
or \(A=22.7^{\circ}, a=3.9\)
Figure \(\PageIndex{9}\)
Example \(\PageIndex{5}\)
Solve the triangle in which \(C=73^{\circ}\), \(a=15\), and \(c=2\). Sketch your result.
Solution
Building a table, we get the following:\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline A = ? & a = 15 \\[6pt] \hline B = ? & b = ? \\[6pt] \hline C = 73^{\circ} & c = 2 \checkmark \\[6pt] \hline \end{array}\nonumber\]Seeing that we have a complete row, we use that to help us find the missing angle from the first row (angle \(A\)).\[\begin{array}{rrclcl} & \dfrac{\sin\left(A\right)}{a} & = & \dfrac{\sin\left(C\right)}{c} & \quad & \left(\text{Law of Sines}\right) \\[6pt] \implies & \sin\left(A\right) & = & \dfrac{a\,\sin\left(C\right)}{c} & \quad & \left(\text{multiplying both sides by }a\right) \\[6pt] \implies & \widehat{A} & = & \sin^{-1}\left(\dfrac{a\,\sin\left(C\right)}{c}\right) & \quad & \left(\text{finding the reference angle}\right) \\[6pt] \end{array}\nonumber\]Grabbing a calculator, substituting in the known values of \(C\), \(a\), and \(c\), we find the calculator returns an error!
What happened?
Consider the following value:\[\dfrac{a\,\sin\left(C\right)}{c} = \dfrac{15\sin\left(73^{\circ}\right)}{2} \approx 7.2\nonumber\]This is what \(\sin\left(A\right)\) is supposed to equal; however, we know that the sine function is always between \(-1\) and \(1\).
What does this mean?
This means there is no such triangle where \(C=73^{\circ}\), \(a=15\), and \(c=2\). We can see this from a sketch.
Figure \(\PageIndex{10}\)
The angle \(C\) is large, but the length of its opposite side is relatively small (when compared to side length \(a\)). If we imagine side \(c\) being able to swing freely, we can see it will never intersect side \(b\). Therefore, forming a triangle with the given information is impossible.
Example \(\PageIndex{5}\) illustrates the final possibility when using the Law of Sines to find a missing angle. In summary, when using the Law of Sines to find a missing angle within a triangle, one of three possibilities occurs:
There is a single triangle satisfying the given information. In this case, the angle will directly result from the arcsine when using the Law of Sines.
Two triangles satisfy the given information. In this case, one triangle will be the direct result of the arcsine, and the second triangle will contain the supplement of that angle.
There is no such triangle satisfying the given information. In this case, your calculator will return a domain error when computing the arcsine.
Checkpoint \(\PageIndex{5}\)
You are told that, in a triangle, \(B = 30^{\circ}\), \(b=7\), and \(c = 15\). Without a calculator, explain how you know no such triangle satisfies the given information.
Answer
While using the Law of Sines to find a missing angle in a triangle, you must compute\[\sin^{-1}\left(\dfrac{15\sin\left(30^{\circ}\right)}{7}\right) = \sin^{-1}\left(\dfrac{15}{14}\right).\nonumber\]This means that \(\sin\left(C\right) = \frac{15}{14}\), which is not possible.
Visualizing the Law of Sines Cases
Up to this point, we have computationally justified that the Law of Sines results in one, two, or no triangles for a given set of information; however, the visual of why this is true is often more impactful.
If we know two sides \(a\) and \(b\) of a triangle and the acute angle \(\alpha\) opposite one of them, there may be one solution, two solutions, or no solution, depending on the size of \(a\) in relation to \(b\) and \(\alpha\), as shown below.
The Law of Sines - Ambiguous Case
In each of the following cases, \( h = b \sin\left( \alpha \right) \) is the altitude of the triangle. Therefore, \( b \gt h \).
The "No Solution" Case
Condition: \(a \lt h\)
Reason: \(a\) is shorter than the altitude of the (non-existent) triangle. Therefore, it is impossible for side \( a \) to intersect the base of the "triangle."
The "One Solution" Cases
Condition #1: \(a = h\)
Reason: \(a\) is exactly the right length to make a right triangle.
Condition #2: \( a \gt b \gt h \)
Reason: If \( a \gt b \), then one triangle is created as side \( a \) swings downward to intersect the base; however, as that side continues to swing, it is too long to strike the base again. Hence, there can be only one triangle.
The "Two Solutions" Case
Condition: \(b \gt a \gt h\)
Reason: Since \( a \gt h \), side \( a \) will strike the base a it swings downward, but before becoming vertical. Moreover, because \( a \lt b \), side \( a \) will strike the base once more as it continues to swing clockwise.
Applications Requiring the Law of Sines
The Law of Sines is used quite a bit in applications within Trigonometry. While I could not cover all possible applications, I have chosen two that commonly occur.
Example \(\PageIndex{6}\)
A vertical communications tower is located at the top of a steep hill, as shown below. The angle of inclination of the hill is \(56^{\circ}\). A guy wire is to be attached to the top of the tower and the ground 175 yards downhill from the base of the tower. The angle formed by the guy wire is \(26^{\circ}\). Find the length of the cable required for the guy wire. Round your answer to the nearest yard.
Figure \(\PageIndex{11}\)
Solution
While it's nice that we have been given a sketch of the situation, I find it's best to redraw the sketch without the "fluff," labeling vertices in uppercase and side lengths in lowercase (as we have been doing this entire time).
Figure \(\PageIndex{12}\)
We now build our table.\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline A = 26^{\circ} & a = ? \\[6pt] \hline B = ? & b = ? \\[6pt] \hline C = ? & c = 175 \\[6pt] \hline \end{array}\nonumber\]We do not have a complete row in this table, so we might think the Law of Sines is not going to be useable; however, looking at the provided sketch, we see one additional piece of information that we have not used.
The green line (side \(c\)) makes an angle of \(56^{\circ}\) with the horizontal. If we draw a horizontal line at vertex \(B\), we can label the angle between the horizontal and the extended green line as \(56^{\circ}\) (see Figure \(\PageIndex{13}\) below). Since the communications tower (side \(a\)) is vertical, the angle between the extended green line and the tower is the complement of \(56^{\circ}\) (again, see Figure \(\PageIndex{13}\) below). This is \(34^{\circ}\).
Figure \(\PageIndex{13}\)
Finally, we can see that the angle within the triangle at vertex \(B\) is the supplement of this \(34^{\circ}\) angle. Hence, \(B = 180^{\circ} - 34^{\circ} = 146^{\circ}\). Let's go ahead and place this information in our table.\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline A = 26^{\circ} & a = ? \\[6pt] \hline B = 146^{\circ} & b = ? \\[6pt] \hline C = ? & c = 175 \\[6pt] \hline \end{array}\nonumber\]Again, we do not have a complete row; however, now that we have two angles, we can easily find the third.\[C = 180^{\circ} - A - B = 8^{\circ}\nonumber\]Filling in the table with this information, we get the following:\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline A = 26^{\circ} & a = ? \\[6pt] \hline B = 146^{\circ} & b = ? \\[6pt] \hline C = 8^{\circ} & c = 175\,\checkmark \\[6pt] \hline \end{array}\nonumber\]Finally, we have a complete row in the table, which tells us that the Law of Sines is now available. We are being asked to find side length \(b\). Therefore,\[\begin{array}{rrclcl} & \dfrac{b}{\sin\left(B\right)} & = & \dfrac{c}{\sin\left(C\right)} & \quad & \left(\text{Law of Sines}\right) \\[6pt] \implies & b & = & \dfrac{c\,\sin\left(B\right)}{\sin\left(C\right)} & \quad & \left(\text{multiplying both sides by }\sin\left(B\right)\right) \\[6pt] \end{array}\nonumber\]Substituting in the values of \(B\), \(C\), and \(c\), we get\[b = \dfrac{175\sin\left(146^{\circ}\right)}{\sin\left(8^{\circ}\right)} \approx 703.\nonumber\]Hence, the length of the guy wire is approximately 703 yards.
Checkpoint \(\PageIndex{6}\)
Lap and Piyali are on the north side of a river that runs east to west. Lap is standing 225 meters due west of Piyali. They spot a treasure chest on the south shore of the river. For Lap, the angle between the chest and Piyali is \(81^{\circ}\), and for Piyali, the angle between the chest and Lap is \(58^{\circ}\) (see Figure \(\PageIndex{14}\) below). How far is Lap from the chest? Round your answer to the nearest meter.
Figure \(\PageIndex{14}\)
Answers
291 meters.
The following example (and the subsequent Checkpoint) illustrates that the Law of Sines can be used when you might otherwise use Right Triangle Trigonometry.
Example \(\PageIndex{7}\)
Ron wants to measure the height of a castle controlled by hostile forces. When he is as close as he can get to the castle, the angle of elevation to the top of the wall is \(18.5^{\circ}\). He then retreats 20 yards and measures the angle of elevation again; this time, it is \(15.9^{\circ}\). How tall is the castle?
Figure \(\PageIndex{15}\)
Solution
Notice that \(h\) is one side of the right triangle \(\triangle ADC\). If we find its hypotenuse, labeled \(r\) in Figure \(\PageIndex{15}\), we can use the sine ratio to find \(h\). To find \(r\), we consider a second triangle, \(\triangle ABC\), as shown in Figure \(\PageIndex{16}\).
Figure \(\PageIndex{16}\)
In this triangle, we know side \(BC = 20\) and would like to find side \(AC = r\). We can use the Law of Sines to find \(r\), but first, we must calculate the other angles of the triangle.
Now, the angle opposite \(r\), \(\angle ABC\), is the complement of \(18.5^{\circ}\), so\[\angle ABC = 180^{\circ} - 18.5^{\circ} = 161.5^{\circ}\nonumber\]The angle opposite the 20-yard side, \(\angle BAC\), is\[\angle BAC=180^{\circ}-\left(161.5^{\circ}+15.9^{\circ}\right)=2.6^{\circ}\nonumber\]Now we can apply the Law of Sines to find \(r\). We have\[\begin{array}{rrclcl} & \dfrac{r}{\sin\left(161.5^{\circ}\right)} & = & \dfrac{20}{\sin\left(2.6^{\circ}\right)} & \quad & \left(\text{Law of Sines}\right) \\[6pt] \implies & r & = & \dfrac{20\sin\left(161.5^{\circ}\right)}{\sin\left(2.6^{\circ}\right)} & \quad & \left(\text{multiplying both sides by }\sin\left(161.5^{\circ}\right)\right) \\[6pt] \end{array}\nonumber\]Finally, using the right triangle \(\triangle ADC\), we can write\[\begin{array}{rrclcl} & \dfrac{h}{r} & = & \sin\left(15.9^{\circ}\right) & \quad & \left(\text{right triangle definition of the sine}\right) \\[6pt] \implies & h & = & r\,\sin\left(15.9^{\circ}\right) & \quad & \left(\text{multiplying both sides by }r\right) \\[6pt] & & = & \dfrac{20\sin\left(161.5^{\circ}\right)}{\sin\left(2.6^{\circ}\right)}\cdot\sin\left(15.9^{\circ}\right) & & \\[6pt] & & \approx & 38.33 & & \\[6pt] \end{array}\nonumber\]The castle is about \(38.33\) yards tall.
Checkpoint \(\PageIndex{7}\)
Solve the problem in the previous example again, but instead of finding \(r\), find the length \(AB\), and then use \(\triangle ABD\) to find \(h\).
Answer
\(AB=120.79\), the castle is about \(38.33\) yards tall.
If you look at a nearby object and alternately close your left and right eyes, the object seems to jump in position. This apparent change occurs because your eyes view the object from two positions spaced several centimeters apart. If the object at point \(O\) is straight ahead of one eye, it appears to be at some angle \(p\) away from the line of sight of the other eye. The angle \(p\) is called the parallax of the object.
Use Figure \(\PageIndex{17}\) to see that \(p\) is also the angle between the directions to your two eyes when viewed from point \(O\). (What fact from Geometry justifies this statement?)
Figure \(\PageIndex{17}\)
Astronomers use parallax to determine the distance from Earth to stars and other celestial objects. Two observers on Earth at a known distance apart both measure the direction to the star. The difference in angle between those two directions is the parallax.
Example \(\PageIndex{8}\)
Astronomers 1000 kilometers apart observe an asteroid with a parallax of \(0.001^{\circ}\). How far is the asteroid from Earth?
Solution
We let \(x\) represent the distance to the asteroid. The asteroid and the two observers make an isosceles triangle with a base of approximately \(1000\) km and equal sides of length \(x\), as shown below.
Figure \(\PageIndex{18}\)
The base angles of the triangle are both \(\frac{180^{\circ}-0.001^{\circ}}{2}=89.999^{\circ}\). Thus,\[\begin{array}{rrclcl} & \dfrac{x}{\sin\left(89.999^{\circ}\right)} & = & \dfrac{1000}{\sin\left(0.001^{\circ}\right)} & \quad & \left(\text{Law of Sines}\right) \\[6pt] \implies & x & = & \dfrac{1000\sin\left(89.999^{\circ}\right)}{\sin\left(0.001^{\circ}\right)} & \quad & \left(\text{multiplying both sides by }\sin\left(89.999^{\circ}\right)\right) \\[6pt] \implies & x & \approx & 57,000,000 & & \\[6pt] \end{array}\nonumber\]The asteroid is about 57 million kilometers from Earth (roughly one-third of the distance to the Sun).
Checkpoint \(\PageIndex{8}\)
Two observers 800 kilometers apart observe an object with a parallax of \(0.0005^{\circ}\). How far is the object from Earth?
_Example.png?revision=1&size=bestfit&width=667&height=550)
