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Law of Cosines

- Calculus
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Let me paint you a picture. Two students, Sam and Monica are comparing how long each of them takes to cycle to school from their homes. They have deduced that Sam's home is 4 miles away from school while Monica's is 3 miles away. They have also found out that the angle formed by the two distances from their homes to the school is 43^{o}. Let us sketch this out. We find that a triangle is constructed as shown below.

Real-world example 1, Aishah Amri - StudySmarter Originals

Now, from this information, is there some sort of method by which we can determine the distance between Sam and Monica's house? Fortunately, you are in luck! There is indeed a formula that can help us solve this problem. We can use the Law of Cosines. In this topic, we shall be introduced to the Law of Cosines and its application to solving triangles.

The Law of Cosines is defined by the rule

${a}^{2}={b}^{2}+{c}^{2}-2bc\mathrm{cos}A$.

Before we begin, let us ease ourselves into this topic by recalling the Cosine Ratio and Pythagoras' Theorem. That way we can bridge the two concepts and derive the Law of Cosines as mentioned above.

Let us first recall the Pythagoras Theorem for a given right-angle triangle with an angle θ.

Right-angle triangle, Aishah Amri - StudySmarter Originals

Thus is given by,

${c}^{2}={a}^{2}+{b}^{2}$.

Furthermore, recall that the cosine of an angle is found by dividing the adjacent by the hypotenuse. This is known as the Cosine Ratio

$\mathrm{cos}\theta =\frac{adjacent}{hypotenuse}=\frac{b}{c}$.

Now that we have set the ideas above, let us use them as a building block in establishing the Law of Cosines. Consider the triangle ABC below.

Law of Cosines, Aishah Amri - StudySmarter Originals

The triangle ABC is split into two triangles: ABD and BCD. Both these triangles are right triangles divided by the perpendicular leg BD of measure h. The side b is also divided into two sections AD = x and CD = b – x. Furthermore, the hypotenuse of triangle ABD is AB = c while the hypotenuse of triangle BCD is BC = a.

Let us analyse the relationship between the lengths of a, b, c and the angle A. By Pythagoras Theorem, we can solve triangle DBC by

${a}^{2}={(b-x)}^{2}+{h}^{2}$.Expanding (b - x)^{2}, we obtain

${a}^{2}={{b}^{2}-2bx+x}^{2}+{h}^{2}$.

Similarly, we can do this for triangle ADB as

${c}^{2}={x}^{2}+{h}^{2}$.

Substituting this expression for x^{2} + h^{2} into the previous equation, we obtain

${a}^{2}={{b}^{2}-2bx+c}^{2}$.

Notice that we can further write x in the form of a Cosine Ratio as

$\mathrm{cos}A=\frac{x}{c}\Rightarrow x=c\mathrm{cos}A$.

Replacing this expression for x, we obtain

${a}^{2}={b}^{2}-2b\times c\mathrm{cos}A+{c}^{2}$.

Applying the Commutative Property and rearranging this, we obtain the Law of Cosines

${a}^{2}={b}^{2}+{c}^{2}-2bc\mathrm{cos}A$.

as we have stated at the beginning of this section.

The Law of Cosines has two other variations from the one given above. By the Law of Cosines, we see that the formula gives us the measure of one unknown side of a triangle. However, a triangle has three sides and three angles. In some cases, we may need to apply this concept on any of the other sides should we require to find their measures. Let us consider the triangle ABC below.

Variations of the Law of Cosines, Aishah Amri - StudySmarter Originals

Here, the value of a, b and c represent the length of each side of this triangle. Angles A, B and C describe the opposite angles for each of these sides respectively. There are three forms of the Law of Cosines for this triangle.

${a}^{2}={b}^{2}+{c}^{2}-2bc\mathrm{cos}A\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{b}^{2}={a}^{2}+{c}^{2}-2ac\mathrm{cos}B\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{c}^{2}={a}^{2}+{b}^{2}-2ab\mathrm{cos}C$

The forms above are suitable for finding the length of an unknown side given the measures of two sides and their included angle. However, if we are given the values of three sides, we may need to perform some lengthy algebra to determine the unknown angle. To simplify such calculations, we can rearrange the expressions above so that we get an explicit formula for the unknown angle.

$\mathrm{cos}A=\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\phantom{\rule{0ex}{0ex}}\mathrm{cos}B=\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}\phantom{\rule{0ex}{0ex}}\mathrm{cos}C=\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}$

In this section, we shall observe several worked examples that apply the Law of Cosines. We can apply the Law of Cosines for any triangle given the measures of two cases:

The value of two sides and their included angle

The value of three sides

Solve a triangle ABC given b = 17, c = 16 and A = 83^{o}.

**Solution **

Let us make a sketch of this triangle.

Example 1, Aishah Amri - StudySmarter Originals

We begin by using the Law of Cosines to find the length of a.

${a}^{2}={b}^{2}+{c}^{2}-2ac\mathrm{cos}A\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}={\left(17\right)}^{2}+{\left(16\right)}^{2}-2\left(17\right)\left(16\right)\mathrm{cos}\left(83\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=545-544\mathrm{cos}\left(83\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow a=\sqrt{545-544\mathrm{cos}\left(83\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow a\approx 21.88(correcttotwodecimalplaces)$

Thus, a is approximately 21.88 units. From here, we shall use the Law of Sines to find angle C.

$\frac{\mathrm{sin}A}{a}=\frac{\mathrm{sin}C}{c}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{sin}\left(83\right)}{21.88}=\frac{\mathrm{sin}C}{16}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin}C=\frac{16\mathrm{sin}\left(83\right)}{21.88}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow C={\mathrm{sin}}^{-1}\left(\frac{16\mathrm{sin}\left(83\right)}{21.88}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow C\approx 46.{54}^{o}(correcttotwodecimalplaces)$

Thus, C is approximately 46.54^{o}. Finally, we can find angle B by

$A+B+C={180}^{o}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow B={180}^{o}-A-C\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow B\approx {180}^{o}-{83}^{o}-46.{54}^{o}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow B\approx 50.{46}^{o}$

Thus, B is approximately 50.46^{o}.

Solve the triangle below given a = 14, b = 11 and c = 5.

**Solution**

We begin by using the Law of Cosines to find angle A.

$\mathrm{cos}A=\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}A=\frac{{\left(11\right)}^{2}+{\left(5\right)}^{2}-{\left(14\right)}^{2}}{2\left(11\right)\left(5\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}A=-\frac{50}{110}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}A=-\frac{5}{11}\phantom{\rule{0ex}{0ex}}\Rightarrow A={\mathrm{cos}}^{-1}\left(-\frac{5}{11}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow A\approx 117.{04}^{o}(correcttotwodecimalplaces)$

Thus, A is approximately 117.04^{o}. From here, we shall use the Law of Sines to find angle B.

$\frac{\mathrm{sin}A}{a}=\frac{\mathrm{sin}B}{b}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{sin}(117.04)}{14}=\frac{\mathrm{sin}B}{11}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin}B=\frac{11\mathrm{sin}(117.04)}{14}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow B={\mathrm{sin}}^{-1}\left(\frac{11\mathrm{sin}(117.04)}{14}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow B\approx 44.{41}^{o}(correcttotwodecimalplaces)$

Thus, B is approximately 44.41^{o}. Finally, we can find angle C by

$A+B+C={180}^{o}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow C={180}^{o}-A-B\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow C\approx {180}^{o}-117.{04}^{o}-44.{41}^{o}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow C\approx 18.{55}^{o}$

Thus, C is approximately 18.55^{o}.

Randy is training for a marathon. He starts his journey by running 9 miles in one direction. Then, he turns and runs another 11 miles. The two legs of his run form an angle of 79^{o}. How far is Randy from his starting point at the end of the 11-mile leg of his run?

**Solution**

Let us begin by sketching the outline of this problem. The variable d represents the distance from Randy's starting point and the end of his 11-mile leg run.

Real-world example 2, Aishah Amri - StudySmarter Originals

Here, we are given the lengths of two sides and their included angle. Thus, we can use the Law of Cosines to find d.

${d}^{2}={\left(9\right)}^{2}+{\left(11\right)}^{2}-2\left(9\right)\left(11\right)\mathrm{cos}\left(79\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{2}=202-198\mathrm{cos}\left(79\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow d=\sqrt{202-198\mathrm{cos}\left(79\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow d\approx 12.81miles(correcttotwodecimalplaces)$

Therefore, d is approximately 12.81 miles.

Find the step angle X made by the hindfoot of a person whose pace averages 27 inches and stride averages 32 inches.

**Solution**

Let us begin by sketching the outline of this problem. The green dot represents each step made by the hindfoot of a person.

Real-world example 3, Aishah Amri - StudySmarter Originals

Here, we are given the lengths of three sides. Thus, we can use the Law of Cosines to find the step angle, X.

$\mathrm{cos}X=\frac{{\left(32\right)}^{2}+{\left(27\right)}^{2}-{\left(27\right)}^{2}}{2\left(32\right)\left(27\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}X=\frac{1024}{1728}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}X=\frac{16}{27}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow X={\mathrm{cos}}^{-1}\left(\frac{16}{27}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow X\approx 53.{66}^{o}(correcttotwodecimalplaces)$

Therefore, X is approximately 53.66^{o}.

An oblique triangle is one that has no right angle. Below are some examples of oblique triangles.

Oblique triangles, Aishah Amri - StudySmarter Originals

To solve such triangles, we need the length of at least one side and the value of any other two components of a given triangle. If the triangle has a solution, then we must choose whether to begin our solution by applying the Law of Sines or the Law of Cosines. The table below helps us decide on such situations.

Given Components | Method to Begin With |

Two angles and any side | |

Two sides and an angle opposite one of them | |

Two sides and their included angle | Law of Cosines |

Three sides | Law of Cosines |

- The Law of Cosines states that

${a}^{2}={b}^{2}+{c}^{2}-2bc\mathrm{cos}A\phantom{\rule{0ex}{0ex}}{b}^{2}={a}^{2}+{c}^{2}-2ac\mathrm{cos}B\phantom{\rule{0ex}{0ex}}{c}^{2}={a}^{2}+{b}^{2}-2ab\mathrm{cos}C$

- Similarly, this can be written in the forms

$\mathrm{cos}A=\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\phantom{\rule{0ex}{0ex}}\mathrm{cos}B=\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}\phantom{\rule{0ex}{0ex}}\mathrm{cos}C=\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}$

- We can use the Law of Cosines to solve triangles given
The value of two sides and their included angle

The value of three sides

Given a triangle ABC, the law of cosines states that a^{2} = b^{2} + c^{2 }- 2bc Cos A

The formula for the law of cosines is a^{2} = b^{2} + c^{2 }- 2bc Cos A

The law of cosines can be used to measure distance and angles of elevation

You can prove the law of cosines by Pythagoras Theorem

More about Law of Cosines

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