# 5.1: The Law of Cosines

Difficulty Level: At Grade Created by: CK-12

## Introduction

This chapter takes concepts that had only been applied to right triangles and interprets them so that they can be used for any type of triangle. First, the laws of sines and cosines take the Pythagorean Theorem and ratios and apply them to any triangle. The second half of the chapter introduces and manipulates vectors. Vectors can be added, subtracted, multiplied and divided.

## Learning Objectives

• Understand how the Law of Cosines is derived.
• Apply the Law of Cosines when you know two sides and the included angle of an oblique (non-right) triangle (SAS).
• Apply the Law of Cosines when you know all three sides of an oblique triangle.
• Identify accurate drawings of oblique triangles.
• Use the Law of Cosines in real-world and applied problems.

## Derive the Law of Cosines

\begin{align*}\triangle{ABC}\end{align*} contains an altitude \begin{align*}BD\end{align*} that extends from \begin{align*}B\end{align*} and intersects \begin{align*}AC\end{align*}. We will refer to the length of \begin{align*}BD\end{align*} as \begin{align*}y\end{align*}. The sides of \begin{align*}\triangle{ABC}\end{align*} measure \begin{align*}a\end{align*} units, \begin{align*}b\end{align*} units, and \begin{align*}c\end{align*} units. If \begin{align*}DC\end{align*} is \begin{align*}x\end{align*} units long, then \begin{align*}AD\end{align*} measures \begin{align*}(b - x)\end{align*} units.

Using the Pythagorean Theorem we know that:

\begin{align*}& c^2 = y^2 + (b - x)^2 && \text{Pythagorean Theorem} \\ & c^2 = y^2 + b^2 - 2bx + x^2 && \text{Expand}\ (b - x)^2 \\ & c^2 = a^2 + b^2 - 2bx && a^2 = y^2 + x^2\ \text{by Pythagorean Theorem} \\ & c^2 = a^2 + b^2 - 2b(a \cos C) && \cos C = \frac{x}{a}, \ \text{so a} \cos C = x\ \text{(cross multiply)} \\ & c^2 = a^2 + b^2 - 2ab \cos C && \text{Simplify}\end{align*}

We can use a similar process to derive all three forms of the Law of Cosines:

\begin{align*}& a^2 = b^2 + c^2 - 2bc \cos A \\ & b^2 = a^2 + c^2 - 2ac \cos B \\ & c^2 = a^2 + b^2 - 2ab \cos C\end{align*}

Note that if either \begin{align*}\angle{A}, \angle{B}\end{align*} or \begin{align*}\angle{C}\end{align*} is \begin{align*}90^\circ\end{align*} then \begin{align*}\cos 90^\circ = 0\end{align*} and the Law of Cosines is identical to the Pythagorean Theorem.

The Law of Cosines is one tool we use in certain situations involving all triangles: right, obtuse, and acute. It is a general statement relating the lengths of the sides of any general triangle to the cosine of one of its angles. There are two situations in which we can and want to use the Law of Cosines:

1. When we know two sides and the included angle in an oblique triangle and want to find the third side (SAS).
2. When we know all three sides in an oblique triangle and want to find one of the angles (SSS).

## Case #1: Finding the Side of an Oblique Triangle

One case where we can use the Law of Cosines is when we know two sides and the included angle in a triangle (SAS) and want to find the third side.

Example 1: Using \begin{align*}\triangle{DEF}, \angle{E} = 12^\circ, d = 18\end{align*}, and \begin{align*}f = 16.8\end{align*}. Find \begin{align*}e\end{align*}.

Solution: Since \begin{align*}\triangle{DEF}\end{align*} isn’t a right triangle, we cannot use the Pythagorean Theorem or trigonometry functions to find the third side. However, we can use our newly derived Law of Cosines.

\begin{align*}e^2 & = 18^2 + 16.8^2 - 2(18)(16.8) \cos 12 && \text{Law of Cosines} \\ e^2 & = 324 + 282.24 - 2(18)(16.8) \cos 12 && \text{Simplify squares} \\ e^2 & = 324 + 282.24 - 591.5836689 && \text{Multiply} \\ e^2 & = 14.6563311 && \text{Add and subtract from left to right} \\ e & \approx 3.8 && \text{Square root}\end{align*}

\begin{align*}^*\text{Note that the negative answer is thrown out as having no geometric meaning in this case.}\end{align*}

Example 2: An architect is designing a kitchen for a client. When designing a kitchen, the architect must pay special attention to the placement of the stove, sink, and refrigerator. In order for a kitchen to be utilized effectively, these three amenities must form a triangle with each other. This is known as the “work triangle.” By design, the three parts of the work triangle must be no less than 3 feet apart and no more than 7 feet apart. Based on the dimensions of the current kitchen, the architect has determined that the sink will be 3.6 feet away from the stove and 5.7 feet away from the refrigerator. If the sink forms a \begin{align*}103^\circ\end{align*} angle with the stove and the refrigerator, will the distance between the stove and the refrigerator remain within the confines of the work triangle?

Solution: In order to find the distance from the sink to the refrigerator, we need to find side \begin{align*}x\end{align*}. To find side \begin{align*}x\end{align*}, we will use the Law of Cosines because we are dealing with an obtuse triangle (and thus have no right angles to work with). We know the length two sides: the sink to the stove and the sink to the refrigerator. We also know the included angle (the angle between the two known lengths) is \begin{align*}103^\circ\end{align*}. This means we have the SAS case and can apply the Law of Cosines.

\begin{align*}x^2 & = 3.6^2 + 5.7^2 - 2(3.6)(5.7) \cos 103 && \text{Law of Cosines} \\ x^2 & = 12.96 + 32.49 - 2(3.6)(5.7) \cos 103 && \text{Simplify squares} \\ x^2 & = 12.96 + 32.49 + 9.23199127 && \text{Multiply} \\ x^2 & = 54.68199127 && \text{Evaluate} \\ x & \approx 7.4 && \text{Square root} \end{align*}

No, this triangle does not conform to the definition of a work triangle. The sink and the refrigerator are too far apart by 0.4 feet.

## Case #2: Finding any Angle of a Triangle

Another situation where we can apply the Law of Cosines is when we know all three sides in a triangle (SSS) and we need to find one of the angles. The Law of Cosines allows us to find any of the three angles in the triangle. First, we will look at how to apply the Law of Cosines in this case, and then we will look at a real-world application.

Example 3: Continuing on from Example 2, if the architect moves the stove so that it is 4.2 feet from the sink and makes the fridge 6.8 feet from the stove, how does this affect the angle the sink forms with the stove and the refrigerator?

Solution: In order to find how the angle is affected, we will again need to utilize the Law of Cosines, but because we do not know the measures of any of the angles, we solve for \begin{align*}Y\end{align*}.

\begin{align*}6.8^2 & = 4.2^2 + 5.7^2 - 2(4.2)(5.7) \cos Y && \text{Law of Cosines} \\ 46.24 & = 17.64 + 32.49 - 2(4.2)(5.7) \cos Y && \text{Simplify squares} \\ 46.24 & = 17.64 + 32.49 - 47.88 \cos Y && \text{Multiply} \\ 46.24 & = 50.13 - 47.88 \cos Y && \text{Add} \\ - 3.89 & = -47.88 \cos Y && \text{Subtract} \\ 0.0812447786 & = \cos Y && \text{Divide} \\ 85.3^\circ & \approx Y && \cos^{-1} \ (0.081244786)\end{align*}

The new angle would be \begin{align*}85.3^\circ\end{align*}, which means it would be \begin{align*}17.7^\circ\end{align*} less than the original angle.

Example 4: In oblique \begin{align*}\triangle{MNO}, m = 45, n = 28\end{align*}, and \begin{align*}o = 49\end{align*}. Find \begin{align*}\angle{M}\end{align*}.

Solution: Since we know all three sides of the triangle, we can use the Law of Cosines to find \begin{align*}\angle{M}\end{align*}.

\begin{align*}45^2 & = 28^2 + 49^2 - 2(28)(49) \cos M && \text{Law of Cosines} \\ 2025 & = 784 + 2401 - 2(28)(49) \cos M && \text{Simplify squares} \\ 2025 & = 784 + 2401 - 2744 \cos M && \text{Multiply} \\ 2025 & = 3185 - 2744 \cos M && \text{Add} \\ -1160 & = -2744 \cos M && \text{Subtract}\ 3185 \\ 0.422740525 & = \cos M && \text{Divide by}\ -2744 \\ 65^\circ & \approx M && \cos^{-1} \ (0.422740525) \end{align*}

It is important to note that we could use the Law of Cosines to find \begin{align*}\angle{N}\end{align*} or \begin{align*}\angle{O}\end{align*} also.

Example 5: Sam is building a retaining wall for a garden that he plans on putting in the back corner of his yard. Due to the placement of some trees, the dimensions of his wall need to be as follows: side \begin{align*}1 = 12ft\end{align*}, side \begin{align*}2 = 18ft\end{align*}, and side \begin{align*}3 = 22ft\end{align*}. At what angle do side 1 and side 2 need to be? Side 2 and side 3? Side 1 and side 3?

Solution: Since we know the measures of all three sides of the retaining wall, we can use the Law of Cosines to find the measures of the angles formed by adjacent walls. We will refer to the angle formed by side 1 and side 2 as \begin{align*}\angle{A}\end{align*}, the angle formed by side 2 and side 3 as \begin{align*}\angle{B}\end{align*}, and the angle formed by side 1 and side 3 as \begin{align*}\angle{C}\end{align*}. First, we will find \begin{align*}\angle{A}\end{align*}.

\begin{align*}22^2 & = 12^2 + 18^2 - 2(12)(18) \cos A && \text{Law of Cosines} \\ 484 & = 144 + 324 - 2(12)(18) \cos A && \text{Simplify squares} \\ 484 & = 144 + 324 - 432 \cos A && \text{Multiply} \\ 484 & = 468 - 432 \cos A && \text{Add} \\ 16 & = -432 \cos A && \text{Subtract}\ 468 \\ -0.037037037 & \approx \cos A && \text{Divide by}\ -432 \\ 92.1^\circ & \approx A && \cos^{-1} \ (-0.037037037) \end{align*}

Next we will find the measure of \begin{align*}\angle{B}\end{align*} also by using the Law of Cosines.

\begin{align*}18^2 & = 12^2 + 22^2 - 2(12)(22) \cos B && \text{Law of Cosines} \\ 324 & = 144 + 484 - 2(12)(22) \cos B && \text{Simplify squares} \\ 324 & = 144 + 484 - 528 \cos B && \text{Multiply} \\ 324 & = 628 - 528 \cos B && \text{Add} \\ -304 & = -528 \cos B && \text{Subtract}\ 628 \\ 0.575757576 & = \cos B && \text{Divide by}\ -528 \\ 54.8^\circ & \approx B && \cos^{-1} \ (0.575757576) \end{align*}

Now that we know two of the angles, we can find the third angle using the Triangle Sum Theorem, \begin{align*}\angle{C} = 180 - (92.1 + 54.8) = 33.1^\circ\end{align*}.

## Identify Accurate Drawings of General Triangles

The Law of Cosines can also be used to verify that drawings of oblique triangles are accurate. In a right triangle, we might use the Pythagorean Theorem to verify that all three sides are the correct length, or we might use trigonometric ratios to verify an angle measurement. However, when dealing with an obtuse or acute triangle, we must rely on the Law of Cosines.

Example 6: In \begin{align*}\triangle{ABC}\end{align*} at the right, \begin{align*}a = 32, b = 20\end{align*}, And \begin{align*}c = 16\end{align*}. Is the drawing accurate if it labels \begin{align*}\angle{C}\end{align*} as \begin{align*}35.2^\circ\end{align*}? If not, what should \begin{align*}\angle{C}\end{align*} measure?

Solution: We will use the Law of Cosines to check whether or not \begin{align*}\angle{C}\end{align*} is \begin{align*}35.2^\circ\end{align*}.

\begin{align*}16^2 & = 20^2 + 32^2 - 2(20)(32) \cos 35.2 && \text{Law of Cosines} \\ 256 & = 400 + 1024 - 2(20)(32) \cos 35.2 && \text{Simply squares} \\ 256 & = 400 + 1024 - 1045.94547 && \text{Multiply} \\ 256 & \neq 378.05453 && \text{Add and subtract}\end{align*}

Since \begin{align*}256 \neq 378.05453\end{align*}, we know that \begin{align*}\angle{C}\end{align*} is not \begin{align*}35.2^\circ\end{align*}. Using the Law of Cosines, we can figure out the correct measurement of \begin{align*}\angle{C}\end{align*}.

\begin{align*}16^2 & = 20^2 + 32^2 -2(20)(32) \cos C && \text{Law of Cosines} \\ 256 & = 400 + 1024 - 2(20)(32) \cos C && \text{Simplify Squares} \\ 256 & = 400 + 1024 - 1280 \cos C && \text{Multiply} \\ 256 & = 1424 - 1280 \cos C && \text{Add} \\ -1168 & = -1280 \cos C && \text{Subtract}\ 1424 \\ 0.9125 & = \cos C && \text{Divide} \\ 24.1^\circ & \approx \angle {C} && \cos^{-1}(0.9125) \end{align*}

For some situations, it will be necessary to utilize not only the Law of Cosines, but also the Pythagorean Theorem and trigonometric ratios to verify that a triangle or quadrilateral has been drawn accurately.

Example 7: A builder received plans for the construction of a second-story addition on a house. The diagram shows how the architect wants the roof framed, while the length of the house is 20 ft. The builder decides to add a perpendicular support beam from the peak of the roof to the base. He estimates that new beam should be 8.3 feet high, but he wants to double-check before he begins construction. Is the builder’s estimate of 8.3 feet for the new beam correct? If not, how far off is he?

Solution: If we knew either \begin{align*}\angle{A}\end{align*} or \begin{align*}\angle{C}\end{align*}, we could use trigonometric ratios to find the height of the support beam. However, neither of these angle measures are given to us. Since we know all three sides of \begin{align*}\triangle{ABC}\end{align*}, we can use the Law of Cosines to find one of these angles. We will find \begin{align*}\angle{A}\end{align*}.

\begin{align*}14^2 & = 12^2 + 20^2 - 2(12)(20) \cos A && \text{Law of Cosines} \\ 196 & = 144 + 400 - 480 \cos A && \text{Simplify} \\ 196 & = 544 - 480 \cos A && \text{Add} \\ -348 & = -480 \cos A && \text{Subtract} \\ 0.725 & = \cos A && \text{Divide} \\ 43.5^\circ & \approx \angle{A} && \cos^{-1} (0.725) \end{align*}

Now that we know \begin{align*}\angle{A}\end{align*}, we can use it to find the length of \begin{align*}BD\end{align*}.

\begin{align*}\sin 43.5 & = \frac{x}{12} \\ 12 \sin 43.5 & = x \\ 8.3 & \approx x\end{align*}

Yes, the builder’s estimate of 8.3 feet for the support beam is accurate.

## Points to Consider

• How is the Pythagorean Theorem a special case of the Law of Cosines?
• In the SAS case, is it possible to use the Law of Cosines to find all missing sides and angles?
• In which cases can we not use the Law of Cosines? Explain.
• Give an example of three side lengths that do not form a triangle.

## Review Questions

1. Using each figure and the given information below, decide which side(s) or angle(s) you could find using the Law of Cosines.
Given Information Figure What can you find?
a. \begin{align*}\angle{A} = 50^\circ, b = 8, c = 11\end{align*}
b. \begin{align*}t = 6, r = 7, i = 11\end{align*}
c. \begin{align*}\angle{L} = 79.5^\circ, m = 22.4, p = 13.17\end{align*}
d. \begin{align*}q = 17, d = 12.8, r = 18.6, \angle{Q} = 62.4^\circ\end{align*}
e. \begin{align*}\angle{B} = 67.2^\circ, d = 43, e = 39\end{align*}
f. \begin{align*}c = 9, d = 11, m = 13\end{align*}
1. Using the figures and information from the chart above, use the Law of Cosines to find the following:
1. side \begin{align*}a\end{align*}
2. the largest angle
3. side \begin{align*}l\end{align*}
4. the smallest angle
5. side \begin{align*}b\end{align*}
6. the second largest angle
2. In \begin{align*}\triangle{CIR}, c = 63, i = 52\end{align*}, and \begin{align*}r = 41.9\end{align*}. Find the measure of all three angles.
3. Find \begin{align*}AD\end{align*} using the Pythagorean Theorem, Law of Cosines, trig functions, or any combination of the three.
4. Find \begin{align*}HK\end{align*} using the Pythagorean Theorem, Law of Cosines, trig functions, or any combination of the three if \begin{align*}JK = 3.6, KI = 5.2, JI = 1.9, HI = 6.7\end{align*}, and \begin{align*}\angle{KJI} = 96.3^\circ\end{align*}.
5. Use the Law of Cosines to determine whether or not the following triangle is drawn accurately. If not, determine how much side d is off by.
6. A businessman is traveling down Interstate 43 and has intermittent cell phone service. There is a transmission tower near Interstate 43. The range of service from the tower forms a \begin{align*}47^\circ\end{align*} angle and the range of service is 26 miles to one section of I-43 and 31 miles to another point on I-43.
1. If the businessman is traveling at a speed of 45 miles per hour, how long will he have service for?
2. If he slows down to 35mph, how much longer will he be able to have service?

7. A dock is being built so that it is 183 yards away from one buoy and 306 yards away from a second buoy. The two buoys are 194.1 yards apart.
1. What angle does the dock form with the two buoys?
2. If the second buoy is moved so that it is 329 yards away from the dock and 207 yards away from the first buoy, how does this affect the angle formed by the dock and the two buoys?

8. A golfer hits the ball from the \begin{align*}18^{th}\end{align*} tee. His shot is a 235 yard hook (curves to the left) \begin{align*}9^\circ\end{align*} from the path straight to the flag on the green.
1. If the tee is 329 yards from the flag, how far is the ball away from the flag?
2. If the golfer’s next shot is 98 yards and is hooked \begin{align*}3^\circ\end{align*} from the path straight to the flag, how far is ball away now?

9. Given the numbers 127, 210 and 17 degrees write a problem that uses the Law of Cosines.
10. The sides of a triangle are 15, 27 and 39. What is its area?
11. A person inherits a piece of land in the shape of a trapezoid as shown, with the side lengths being in feet. What is the area of the piece of land? How many acres is it?

1. side \begin{align*}a\end{align*}
2. \begin{align*}\angle{T}, \angle{R}\end{align*}, and \begin{align*}\angle{I}\end{align*}
3. side \begin{align*}l\end{align*}
4. \begin{align*}\angle{R}\end{align*} and \begin{align*}\angle{D}\end{align*}
5. side \begin{align*}b\end{align*}
6. \begin{align*}\angle{C}, \angle{D}, \angle{M}\end{align*}
1. \begin{align*}a^2 = 8^2 + 11^2 - 2 \cdot 8 \cdot 11 \cdot \cos 50^\circ, a \approx 8.5\end{align*}
2. \begin{align*}11^2 = 6^2 + 7^2 - 2 \cdot 6 \cdot 7 \cdot \cos I, \angle{I} \approx 115.4^\circ\end{align*}
3. \begin{align*}l^2 = 22.4^2 + 13.17^2 - 2 \cdot 22.4 \cdot 13.17 \cdot \cos 79.5^\circ, l \approx 23.8\end{align*}
4. \begin{align*}12.8^2 = 17^2 + 18.6^2 - 2 \cdot 17 \cdot 18.6 \cdot \cos D, \angle{D} \approx 41.8^\circ\end{align*}
5. \begin{align*}b^2 = 39^2 + 43^2 - 2 \cdot 39 \cdot 43 \cdot \cos 67.2^\circ, b \approx 45.5\end{align*}
6. \begin{align*}11^2 = 9^2 + 13^3 - 2 \cdot 9 \cdot 13 \cdot \cos D, \angle{D} \approx 56.5^\circ\end{align*}
1. \begin{align*}63^2 = 52^2 + 41.9^2 - 2 \cdot 52 \cdot 41.9 \cdot \cos C, 52^2 = 63^2 + 41.9^2 - 2 \cdot 63 \cdot 41.9 \cdot \cos I, 180^\circ - 83.5^\circ - 55.1^\circ = 41.4^\circ, \angle{C} \approx 83.5^\circ, \angle{I} \approx 55.1^\circ, \angle{R} \approx 41.4^\circ\end{align*}
2. First, find \begin{align*}AB\end{align*}. \begin{align*}AB^2 = 14.2^2 + 15^2 - 2 \cdot 14.2 \cdot 15 \cdot \cos 37.4^\circ, AB = 9.4. \sin 23.3^\circ = \frac{AD}{9.4}, AD = 3.7.\end{align*}
3. \begin{align*}\angle{HJI} = 180^\circ - 96.3^\circ = 83.7^\circ\end{align*} (these two angles are a linear pair). \begin{align*}6.7^2 = HJ^2 + 1.9^2 - 2 \cdot HJ \cdot 1.9 \cdot \cos 83.7^\circ\end{align*}. This simplifies to the quadratic equation \begin{align*}HJ^2 - 0.417HJ - 41.28\end{align*}. Using the quadratic formula, we can determine that \begin{align*}HJ \approx 6.64\end{align*}. So, since \begin{align*}HJ + JK = HK, 6.64 + 3.6 \approx HK \approx 10.24\end{align*}.
4. To determine this, use the Law of Cosines and solve for \begin{align*}d\end{align*} to determine if the picture is accurate. \begin{align*}d^2 = 12^2 + 24^2 - 2 \cdot 12 \cdot 24 \cdot \cos 30^\circ, d = 14.9\end{align*}, which means \begin{align*}d\end{align*} in the picture is off by 1.9.
5. (a) First, find \begin{align*}x\end{align*}: \begin{align*}x^2 = 31^2 + 26^2 - 2 \cdot 31 \cdot 26 \cdot \cos 47^\circ, x = 23.187\ miles\end{align*}. Dividing the miles by his speed will tell us how long he will have service. \begin{align*}\frac{23.187}{45} = 0.52\ hr\end{align*} or \begin{align*}30.9\ min\end{align*}. (b) \begin{align*}\frac{23.187}{35} = 0.66\ hr\end{align*} or \begin{align*}39.7\ min\end{align*}, so he will have service for 8.8 minutes longer.
1. \begin{align*}194.1^2 = 183^2 + 306^2 - 2 \cdot 183 \cdot 306 \cdot \cos a\end{align*}. The angle formed, \begin{align*}a\end{align*}, is \begin{align*}37^\circ\end{align*}.
2. \begin{align*}207^2 = 183^2 + 329^2 - 2 \cdot 183\cdot 329 \cdot \cos b\end{align*}. The new angle, \begin{align*}b\end{align*}, will need to be \begin{align*}34.8^\circ\end{align*} rather than \begin{align*}37^\circ\end{align*} or \begin{align*}2.2^\circ\end{align*} less.
1. \begin{align*}x^2 = 235^2 + 329^2 - 2 \cdot 235 \cdot 329 \cdot \cos 9^\circ\end{align*}, making the ball 103.6 yards away from the flag.

\begin{align*}x^2 = 98^2 + 103.6^2 - 2 \cdot 98 \cdot 103.6 \cdot \cos 3^\circ\end{align*}, making his second shot 7.7 yards away from the flag.

6. Students answers will vary. The goal is to have each student create their own word problem.
7. We need to find the height in order to get the area. \begin{align*}27^2 & = 15^2 + 39^2 - 2 \cdot 15 \cdot 39 \cdot \cos x, x = 29.6^\circ \\ \sin 29.6^\circ & = \frac{h}{15} \rightarrow h = 7.4\\ A & = \frac{1}{2} \cdot 39 \cdot 7.4 = 144.3\end{align*}
8. Recall that the area of a trapezoid is \begin{align*}A = \frac{1}{2} h (b_1 + b_2)\end{align*}. We need to find the angle \begin{align*}x\end{align*}, in order to find \begin{align*}y\end{align*} and then \begin{align*}h\end{align*}. \begin{align*}2100^2 & = 2400^2 + 2200^2 - 2 \cdot 2400 \cdot 2200 \cdot \cos x, x = 54.1^\circ. \\ 90^\circ - 54.1^\circ & = 35.9^\circ = y. \cos 35.9^\circ = \frac{h}{2200} \rightarrow h = 1782.1 . \\ A & = \frac{1}{2}1782.1 (2400+3000) = 4,811,670\ sq. ft.\ \text{or}\ 110.5\ acres.\end{align*}

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