This section is a very traditional treatment of the Law of Cosines and applications involving the Law of Cosines. You will likely spend a good chunk of time on navigation problems, as these are often difficult for students to master.
The following is a list of learning objectives for this section.
Learning Objectives (click to expand)
Use the Law of Cosines to find a missing side or a missing angle in an oblique triangle.
Use the Law of Sines to find the remaining angle once the Law of Cosines has been used to find one angle in an oblique triangle.
Use the Law of Cosines to solve an application involving heading or bearing.
Use the Law of Cosines to solve an application.
The Law of Cosines
When learning the Law of Sines, we used a table to help organize information and determine which unknown to find first. For example, suppose the information in the following table was what we were given.\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline \angle A = ? & a = ? \\[6pt] \hline \angle B = 79.4^{\circ} & b = ? \\[6pt] \hline \angle C = 17.4^{\circ} & c = 400 \, \checkmark \\[6pt] \hline \end{array} \nonumber\]I mentioned in the previous section that a complete row in this table meant we could use the Law of Sines. Therefore, we can see that, given two angles and a side opposite one of those angles, we can use the Law of Sines to solve the triangle.
Suppose, instead, the given information looked as follows:\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline \angle A = ? & a = ? \\[6pt] \hline \angle B = 79.4^{\circ} & b = 1600 \, \checkmark \\[6pt] \hline \angle C = ? & c = 400 \\[6pt] \hline \end{array} \nonumber\]Again, we have a complete row, so we can use the Law of Sines. Therefore, given two sides and an angle opposite one of those sides, we could use the Law of Sines to solve the triangle (if one exists).
In contrast, given any other combination of sides or angles, the Law of Sines is not helpful. If you consider what is required for this to happen, we just need a table with no complete rows. The two major examples are the following:\[\begin{array}{ccc} \begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline \angle A = ? & a = 7 \\[6pt] \hline \angle B = 115^{\circ} & b = ? \\[6pt] \hline \angle C = ? & c = 5 \\[6pt] \hline \end{array} & \quad \text{or} \quad & \begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline \angle A = ? & a = 7 \\[6pt] \hline \angle B = ? & b = 9 \\[6pt] \hline \angle C = ? & c = 5 \\[6pt] \hline \end{array} \\[6pt] \end{array} \nonumber\]Notice that there is not a complete row within either of these tables and, as of now, we have no simple way of completing a row. Not having an angle-side pairing presents an issue that the Law of Sines cannot surmount (unless you were given two angles, in which case you can find the third and proceed to use the Law of Sines).
To solve a triangle when we know two sides and the included angle, we will need a generalization of the Pythagorean Theorem.
In a right triangle, with \(C = 90^{\circ}\), the Pythagorean Theorem tells us that\[c^2 = a^2 + b^2. \nonumber\]If we allow angle \(C\) to vary (rather than being locked in at a right angle) but keep \(a\) and \(b\) the same length, the side \(c\) will grow or shrink, depending on whether we increase or decrease the angle \(C\), as shown below.
Figure \(\PageIndex{1}\)
The Pythagorean Theorem is a special case of a more general law that applies to all triangles, no matter the size of angle \(C\). This law is known as the Law of Cosines.
Theorem: Law of Cosines
If the angles of a triangle are \(A\), \(B\), and \(C\), and the opposite sides are respectively \(a\), \(b\), and \(c\), then\[ \begin{array}{rcl}
a^2 & = & b^2+c^2-2 b c \cos \left(A\right) \\[6pt] b^2 & = & a^2+c^2-2 a c \cos\left( B\right) \\[6pt] c^2 & = & a^2+b^2-2 a b \cos \left(C\right) \\[6pt] \end{array} \nonumber \]
Proof
Start with general triangles as seen in the figures below.
Two general triangles, one acute (left) and one obtuse (right), with altitudes labeled as \( h \)
We will perform the proof for the acute triangle (the proof associated with the obtuse triangle is similar).
The acute triangle is comprised of two right triangles - \( \triangle ABD \) and \( \triangle BCD \). Let \( x \) be the length of the line segment \( \overline{AD} \). Then the length of the segment \( \overline{CD} \) is \( b - x \). We show these facts in the following figure, splitting the general right triangle into its two right triangle pieces.
"Splitting" the acute triangle into two right triangles
With the leftmost triangle, we have the following:\[ \begin{array}{rrclcl}
& \left( b - x \right)^2 + h^2 & = & a^2 & \quad & \left(\text{Pythagorean Theorem}\right) \\[6pt] \implies & b^2 - 2bx + x^2 + h^2 & = & a^2 & \quad & \left( \text{distributing} \right) \\[6pt] \end{array} \nonumber \]With the rightmost triangle, we have\[ x^2 + h^2 = c^2 \quad \left( \text{Pythagorean Theorem} \right) \nonumber \]Therefore,\[ b^2 - 2bx + x^2 + h^2 = a^2 \quad \text{and} \quad x^2 + h^2 = c^2. \nonumber \]Replace \( x^2 + h^2 \) in the first equation with \( c^2 \). Doing so yields the following:\[ b^2 - 2bx + c^2 = a^2 \implies b^2 + c^2 - 2bx = a^2. \nonumber \]From the rightmost triangle in the figure above, we see that \( \cos\left( A \right) = \frac{x}{c} \), and so \( c \, \cos\left( A \right) = x \). Hence, we have developed the following relationship between the sides of any triangle.\[ \begin{array}{rrclcl}
& b^2 + c^2 - 2bx & = & a^2 & & \\[6pt] \implies & b^2 + c^2 - 2b\left( c \, \cos\left( A \right) \right) & = & a^2 & \quad & \left( \text{substitution} \right) \\[6pt] \implies & b^2 + c^2 - 2bc \, \cos\left( A \right) & = & a^2 & & \\[6pt] \end{array} \nonumber \]When \(A\) is a right angle, \(\cos\left( A \right) = \cos\left( 90^{\circ}\right)=0\), so the equation reduces to the Pythagorean Theorem.
We can write similar equations involving the angles \(B\) or \(C\). In all cases, the angle within the cosine is opposite the side playing the role of the "hypotenuse" (really, a quasi-hypotenuse).
The Law of Cosines is also known as the Generalized Pythagorean Theorem.
Rather than committing all three forms to memory, I find it much easier to think of these as modifications of the Pythagorean Theorem. We have a sum of squares of sides almost equal to the square of a quasi-hypotenuse. The adjustment we make to the sum of squares side is to subtract twice the product of the non-quasi-hypotenuse sides and the cosine of the angle opposite the quasi-hypotenuse.
Now that I have written that out, I can see that it might be confusing to most people, so let me say it symbolically. If \(s_1\) and \(s_2\) are playing the roles of the non-quasi-hypotenuse sides, and \(h\) is playing the role of the quasi-hypotenuse, then\[s_1^2 + s_2^2 - 2s_1 s_2 \cos\left( H \right) = h^2, \nonumber\]where \(H\) is the angle opposite side \(h\).
Finding a Side
Example \(\PageIndex{1}\)
The town of Avery lies 48 miles due east of Baker, and Clio is 34 miles from Baker, in the direction \(35^{\circ}\) west of north. How far is it from Avery to Clio?
Solution
We start with a simple sketch.
Figure \(\PageIndex{2}\)
The angle \(\angle ABC=35^{\circ}+90^{\circ}=125^{\circ}\). Thus, in \(\triangle ABC\) we have the following:\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline \angle A = ? & a = 34 \\[6pt] \hline \angle B = 125^{\circ} & b = ? \\[6pt] \hline \angle C = ? & c = 48 \\[6pt] \hline \end{array} \nonumber\]Since we do not have a complete row, we must use the Law of Cosines. We are interested in knowing the length of side \(b\), so that will be our quasi-hypotenuse.\[\begin{array}{rrclcl} & b^2 & = & a^2 + c^2 - 2ac \cos \left(B\right) & \quad & \left( \text{Law of Cosines} \right) \\[6pt] \implies & b^2 & = & 34^2 + 48^2 - 2(34)(48)\cos \left(125^{\circ}\right) & \quad & \left(\text{substitute}\right) \\[6pt] \implies & b^2 & = & 3460 - 3264\cos \left(125^{\circ}\right) & & \\[6pt] \implies & b^2 & \approx & 5332.153 & & \\[6pt] \implies & b & \approx & 73.02 & & \\[6pt] \end{array} \nonumber\]Therefore, Avery is about 73 miles from Clio.
Caution: Follow the Order of Operations
When simplifying the Law of Cosines, follow the Order of Operations carefully. In Example \(\PageIndex{1}\), the right side of the equation\[b^2=34^2+48^2-2(34)(48) \cos \left(125^{\circ}\right)\nonumber\]has three terms, and simplifies to\[\begin{array}{rrcl} & b^2 & = & 1156+2304-3264 \cos \left(125^{\circ}\right) \\[6pt] \implies & b^2 & \approx & 3460-3264(-0.573567364 \ldots) \\[6pt] \end{array}\nonumber\]Note that 3264 is the coefficient of \(\cos \left(125^{\circ}\right)\), so it would be incorrect to subtract 3264 from 3460. Using a graphing calculator, you can enter the right side of the equation exactly as it is written.
Checkpoint \(\PageIndex{1}\)
In \(\triangle ABC\), \(a=11\), \(c=23\), and \(B=87^{\circ}\). Find \(b\), and round your answer to two decimal places.
Answer
24.97
Finding an Angle
We can also use the Law of Cosines to find an angle when we know all three sides of a triangle. In the following example, pay close attention to the algebraic steps used to solve the equation.
Example \(\PageIndex{2}\)
In the triangle below, \(a = 6\), \(b = 7\), and \(c = 11\). Find angle \(C\).
Figure \(\PageIndex{3}\)
Solution
Our table looks like\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline \angle A = ? & a = 6 \\[6pt] \hline \angle B = ? & b = 7 \\[6pt] \hline \angle C = ? & c = 11 \\[6pt] \hline \end{array} \nonumber\]Since there is no way to get a complete row, we must use the Law of Cosines. Moreover, because we are asked to find \(C\), we choose the version of the Law of Cosines that uses angle \(C\).\[\begin{array}{rrclcl} & c^2 & = & a^2 + b^2 - 2ab\cos\left( C\right) & \quad & \left( \text{Law of Cosines} \right) \\[6pt] \implies & 11^2 & = & 6^2 + 7^2 - 2(6)(7)\cos\left( C\right) & \quad & \left( \text{substitute} \right) \\[6pt] \implies & 121 & = & 36 + 49 - 84\cos\left( C\right) & \quad & \left(\text{apply powers and compute the product}\right) \\[6pt] \implies & 36 & = & -84\cos \left(C\right) & \quad & \left(\text{subtract }85\text{ from both sides} \right) \\[6pt] \implies & -\dfrac{3}{7} & = & \cos \left(C\right) & \quad & \left(\text{divide both sides by }-84\right) \\[6pt] \implies & \widehat{C} & = & \cos^{-1}\left(\dfrac{3}{7}\right) & \quad & \left( \text{finding the reference angle} \right) \\[6pt] \implies & C & \approx & 115.4^{\circ} & \quad & \left( \text{arccosine is either in }\mathrm{QI}\text{ or }\mathrm{QII}\text{.} \right. \\[6pt] & & & & & \left. \text{Since the cosine is negative, }C \in \mathrm{QII}\text{.} \right. \\[6pt] & & & & & \left. \text{Thus, }C = 180^{\circ} - \widehat{C} \right) \\[6pt] \end{array} \nonumber\]Angle \(C\) is roughly \(115.4^{\circ}\).
Checkpoint \(\PageIndex{2}\)
In \(\triangle ABC\), \(a=5.3\), \(b=4.7\), and \(c=6.1\). Find angle \(B\), and round your answer to two decimal places.
Answer
\(48.07^{\circ}\)
Once we have calculated one of the angles in a triangle, we can use either the Law of Sines or the Law of Cosines to find a second angle. Here is how we would use the Law of Sines to find angle \(A\) in Example \(\PageIndex{2}\).\[\begin{array}{rrclcl} & \dfrac{\sin \left(A\right)}{a} & = & \dfrac{\sin \left(C\right)}{c} & \quad & \left( \text{Law of Sines} \right) \\[6pt] \implies & \sin \left(A\right) & = & \dfrac{a \, \sin \left( C \right)}{c} & \quad & \left( \text{multiplying both sides by }a \right) \\[6pt] \implies & \widehat{A} & = & \sin^{-1}\left(\dfrac{a \, \sin \left( C \right)}{c}\right) & \quad & \left( \text{finding the reference angle} \right) \\[6pt] \end{array}\nonumber\]Substituting in the given values, we get \(\widehat{A} \approx 29.5^{\circ}\). Since \(A\) is the shortest side of the triangle, \(A\) must be acute. Therefore, \(A \approx 29.5^{\circ}\). Finally,\[B=180^{\circ}-A-C \approx 35.1^{\circ}.\nonumber\]
Alternatively, we can use the Law of Cosines to find angle \(A\).\[\begin{array}{rrclcl} & a^2 & = & b^2+c^2-2 b c \cos \left(A\right) & \quad & \left(\text{Law of Cosines}\right) \\[6pt] \implies & a^2 - b^2 - c^2 & = & -2bc\cos\left( A \right) & \quad & \left( \text{subtracting }b^2\text{ and }c^2\text{ from both sides} \right) \\[6pt] \implies & \dfrac{a^2 - b^2 - c^2}{-2bc} & = & \cos\left(A\right) & \quad & \left( \text{dividing both sides by }-2bc \right) \\[6pt] \implies & \cos^{-1}\left( \left| \dfrac{a^2 - b^2 - c^2}{-2bc} \right| \right) & = & \widehat{A} & \quad & \left( \text{finding the reference angle} \right) \\[6pt] \end{array} \nonumber\]Substituting the given values into this result, we get the reference angle of \(\widehat{A} \approx 29.5^{\circ}\). If we had computed the value of \(\frac{a^2 - b^2 - c^2}{-2bc}\), we would have found that this fraction is positive. Since the arccosine returns in \(\mathrm{QI}\) or \(\mathrm{QII}\), and, of those two quadrants, the cosine is positive in \(\mathrm{QI}\), our actual angle is \(A \approx 29.5^{\circ}\).
From this conversation about finding a second angle once you have used the Law of Cosines to find an initial angle, I hope you agree that the amount of computational work is minimized when using the Law of Sines. You can choose either method to find a second angle, but I will always choose the Law of Sines over the Law of Cosines.
Advice: Choose Your Angle Wisely!
When you have used the Law of Cosines to find the first missing angle within a triangle, you can use the Law of Sines or the Law of Cosines to find the next unknown angle. If using the Law of Sines, however, always choose to find the angle opposite the smallest remaining side. This is because the angle opposite the smallest remaining side is guaranteed to be acute. Hence, the reference angle returned when using the Law of Sines will be the actual angle!
Applications Involving the Law of Cosines
The applications of the Law of Cosines are countless, more so than the Law of Sines; however, they all somewhat follow the structure of the following example.
Example \(\PageIndex{3}\)
A researcher wishes to determine the width of a vernal pond, as drawn below. From a point \(P\), he finds the distance to the eastern-most point of the pond to be \(950\) feet, while the distance to the western-most point of the pond from \(P\) is \(1000\) feet (obviously, the perspective of the provided image is such that due north is roughly facing the 4 o'clock position). If the angle between the two lines of sight is \(60^{\circ}\), find the width of the pond.
Figure \(\PageIndex{4}\)
Solution
Let \(p\) be the side length opposite angle \(P\), \(a = 1000\), and \(b = 950\). Then our table looks like the following:\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline \angle A = ? & a = 1000 \\[6pt] \hline \angle B = ? & b = 950 \\[6pt] \hline \angle P = 60^{\circ} & p = ? \\[6pt] \hline \end{array} \nonumber\]No complete rows (and no hope of getting a complete row using simple arithmetic) means we are using the Law of Cosines. We want to find length \(p\).\[\begin{array}{rrclcl} & p^2 & = & a^2 + b^2 - 2 a b \cos\left( P \right) & \quad & \left( \text{Law of Cosines} \right) \\[6pt] \implies & p & = & \sqrt{a^2 + b^2 - 2 a b \cos\left( P \right)} & \quad & \left( \text{Extraction of Roots} \right) \\[6pt] \end{array} \nonumber\]Substituting in the given values, we get\[p = \sqrt{950^{2}+1000^{2}-2(950)(1000) \cos \left(60^{\circ}\right)} = \sqrt{952{,}500} \approx 976.\nonumber\]Thus, the pond is roughly 976 feet wide.
Checkpoint \(\PageIndex{3}\)
The hour hand on Kim's antique clock is 4 inches long and the minute hand is 5.5 inches long. Find the distance between the ends of the hands when the clock reads four o'clock. Round your answer to the nearest hundredth of an inch.
Answer
8.26 inches.
Revisiting Navigation Problems
In a previous section, we were introduced to navigation problems (heading and bearing problems). At that time, however, we were only able to work with these styles of problems when they resulted in right triangles. We now have enough Trigonometry "under our belts" to approach these problems in any situation.
Example \(\PageIndex{4}\)
The sailing club leaves the marina on a heading of \(15^{\circ}\) and sails for 18 miles. They then change course, and after traveling for 12 miles on a heading of \(35^{\circ}\), they experience engine trouble and radio for help. The marina sends a speed boat to rescue them. How far should the speed boat go, and on what heading?
Solution
As we did when first introduced to heading problems, we sketch the situation, placing small axes at each pivot point.
Figure \(\PageIndex{5}\)
This results in the following figure.
Figure \(\PageIndex{6}\)
We want to find the distance \(d\) and the angle \(\angle BAC\) shown in the figure. We use alternate interior angles to get further information about the angles within \(\triangle ABC\).
Figure \(\PageIndex{7}\)
This results in the final, clean sketch.
Figure \(\PageIndex{8}\)
In \(\triangle ABC\), we calculated the angle at point \(B\) where the sailing club changed course as follows:\[B = 180^{\circ} - 35^{\circ} + 15^{\circ} = 160^{\circ}.\nonumber\] We now build our table by using the Law of Cosines to find \(b\) and \(\angle A\).\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline \angle A = ? & a = 12 \\[6pt] \hline \angle B = 160^{\circ} & b = ? \\[6pt] \hline \angle C = ? & c = 18 \\[6pt] \hline \end{array} \nonumber\]Since there is not a complete row, and there is no way to get a complete row without Trigonometry, we use the Law of Cosines to find the length of side \(b\) (accidentally labeled as side \(d\) in the figure above).\[\begin{array}{rrclcl} & b^2 & = & a^2 + c^2 - 2ac\cos\left( B\right) & \quad & \left( \text{Law of Cosines} \right) \\[6pt] \implies & b & = & \sqrt{a^2 + c^2 - 2ac\cos\left( B\right)} & \quad & \left( \text{Extraction of Roots} \right) \\[6pt] \end{array}\nonumber\]Substituting in the given values, we get \(b \approx 29.56\) miles. Taking the time to add this to our table, we get the following:\[\begin{array}{|c|c|} \hline \textbf{Angle} & \textbf{Opposite Side} \\[6pt] \hline \angle A = ? & a = 12 \\[6pt] \hline \angle B = 160^{\circ} & b \approx 29.56 \, \checkmark \\[6pt] \hline \angle C = ? & c = 18 \\[6pt] \hline \end{array} \nonumber\]Now that we have a complete row, we can use the Law of Sines to find one of the remaining angles. The advice given previously was to find the angle opposite the smallest remaining side. This would be \(A\) (which is, in fact, the angle we need to find).\[\begin{array}{rrclcl} & \dfrac{\sin\left( A \right)}{a} & = & \dfrac{\sin\left( B \right)}{b} & \quad & \left( \text{Law of Sines} \right) \\[6pt] \implies & \sin\left( A \right) & = & \dfrac{a \, \sin\left( B \right)}{b} & \quad & \left( \text{multiplying both sides by }a \right) \\[6pt] \implies & \widehat{A} & = & \sin^{-1}\left(\dfrac{a \, \sin\left( B \right)}{b}\right) & \quad & \left( \text{finding the reference angle} \right) \\[6pt] \end{array} \nonumber\]Substituting in the nonrounded values for \(a\), \(b\), and \(B\), we get \(\widehat{A} \approx 8.0^{\circ}\). Since \(A\) is opposite the smallest remaining side, \(A\) must be acute. Therefore, \(A \approx 8.0^{\circ}\).
We finish this problem with a bit of extra logic. We know the speed boat must travel \(b \approx 29.56\) miles; however, the heading of the speed boat is not \(8.0^{\circ}\). This is because they must rotate through \(15^{\circ}\) and then an additional \(8.0^{\circ}\) to face the sailing club's boat. Thus, their heading is \(23.0^{\circ}\).
Checkpoint \(\PageIndex{4}\)
Roy wants to fly from Anchorage to Nome, Alaska, a distance of 540 miles on a heading of \(303^{\circ}\). After flying for some time, he discovers that his heading is in error and is actually flying at a heading of \(313^{\circ}\). Roy corrects his flight plan and changes course when he is 200 miles from Anchorage. What is his new heading, and how far is he from Nome?
Answer
\(297.2^{\circ}\), \(344.8\) miles
Success in College: Dealing With Failure (Part 4)
A first-year student fails her math class. She hates math. Her advisor wants her to retake the course the next term. The student wants to wait a year or so before trying again. Discuss the advantages and disadvantages of waiting to retake this math course.
Some colleges restrict the number of times a student can repeat a course. Describe any restrictions on repeating a course at your school.
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