11.3 The Law of Cosines

In Section, we developed the Law of Sines (Theorem ) to enable us to solve triangles in the `Angle-Angle-Side' (AAS), the `Angle-Side-Angle' (ASA) and the ambiguous `Angle-Side-Side' (ASS) cases. In this section, we develop the Law of Cosines which handles solving triangles in the `Side-Angle-Side' (SAS) and `Side-Side-Side' (SSS) cases.¹ We state and prove the theorem below.

Theorem

Law of Cosines: Given a triangle with angle-side opposite pairs $(α, a)$ , $(β, b)$ and $(γ, c)$ , the following equations hold

$a^{2} = b^{2} + c^{2} - 2 b c \cos (α) b^{2} = a^{2} + c^{2} - 2 a c \cos (β) c^{2} = a^{2} + b^{2} - 2 a b \cos (γ)$

or, solving for the cosine in each equation, we have

$\cos (α) = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \cos (β) = \frac{a^{2} + c^{2} - b^{2}}{2 a c} \cos (γ) = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$

To prove the theorem, we consider a generic triangle with the vertex of angle $α$ at the origin with side $b$ positioned along the positive $x$ -axis.

Coordinate-plane figure. — **Figure 11.48**

From this set-up, we immediately find that the coordinates of $A$ and $C$ are $A ⁢ open parenthesis 0, 0 close parenthesis$ and $C ⁢ open parenthesis b, 0 close parenthesis$ . From Theorem, we know that since the point $B (x, y)$ lies on a circle of radius $c$ , the coordinates of $B$ are $B (x, y) = B (c \cos (α), c \sin (α))$ . (This would be true even if $α$ were an obtuse or right angle so although we have drawn the case when $α$ is acute, the following computations hold for any angle $α$ drawn in standard position where $0 less than α less than 180 degrees$ .) We note that the distance between the points $B$ and $C$ is none other than the length of side $a$ . Using the distance formula, Equation, we get

$\begin{matrix} a & = & \sqrt{{(c \cos (α) - b)}^{2} + {(c \sin (α) - 0)}^{2}} \\ a^{2} & = & {(\sqrt{{(c \cos (α) - b)}^{2} + c^{2} \sin^{2} (α)})}^{2} \\ a^{2} & = & {(c \cos (α) - b)}^{2} + c^{2} \sin^{2} (α) \\ a^{2} & = & c^{2} \cos^{2} (α) - 2 b c \cos (α) + b^{2} + c^{2} \sin^{2} (α) \\ a^{2} & = & c^{2} (\cos^{2} (α) + \sin^{2} (α)) + b^{2} - 2 b c \cos (α) \\ a^{2} & = & c^{2} (1) + b^{2} - 2 b c \cos (α) & Since \cos^{2} (α) + \sin^{2} (α) = 1 \\ a^{2} & = & c^{2} + b^{2} - 2 b c \cos (α) \end{matrix}$

The remaining formulas given in Theorem can be shown by simply reorienting the triangle to place a different vertex at the origin. We leave these details to the reader. What's important about $a$ and $α$ in the above proof is that $(α, a)$ is an angle-side opposite pair and $b$ and $c$ are the sides adjacent to $α$ – the same can be said of any other angle-side opposite pair in the triangle. Notice that the proof of the Law of Cosines relies on the distance formula which has its roots in the Pythagorean Theorem. That being said, the Law of Cosines can be thought of as a generalization of the Pythagorean Theorem. If we have a triangle in which $γ equals 90 degrees$ , then $\cos (γ) = \cos (90^{\circ}) = 0$ so we get the familiar relationship $c^{2} = a^{2} + b^{2}$ . What this means is that in the larger mathematical sense, the Law of Cosines and the Pythagorean Theorem amount to pretty much the same thing.²

Example 1

Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle.

$β equals 50 degrees$ , $a equals 7$ units, $c equals 2$ units
$a equals 4$ units, $b equals 7$ units, $c equals 5$ units

We are given the lengths of two sides, $a equals 7$ and $c equals 2$ , and the measure of the included angle, $β equals 50 degrees$ . With no angle-side opposite pair to use, we apply the Law of Cosines. We get $b^{2} = 7^{2} + 2^{2} - 2 (7) (2) \cos (50^{\circ})$ which yields $b = \sqrt{53 - 28 \cos (50^{\circ})} \approx 5.92$ units. In order to determine the measures of the remaining angles $α$ and $γ$ , we are forced to used the derived value for $b$ . There are two ways to proceed at this point. We could use the Law of Cosines again, or, since we have the angle-side opposite pair $(β, b)$ we could use the Law of Sines. The advantage to using the Law of Cosines over the Law of Sines in cases like this is that unlike the sine function, the cosine function distinguishes between acute and obtuse angles. The cosine of an acute is positive, whereas the cosine of an obtuse angle is negative. Since the sine of both acute and obtuse angles are positive, the sine of an angle alone is not enough to determine if the angle in question is acute or obtuse. Since both authors of the textbook prefer the Law of Cosines, we proceed with this method first. When using the Law of Cosines, it's always best to find the measure of the largest unknown angle first, since this will give us the obtuse angle of the triangle if there is one. Since the largest angle is opposite the longest side, we choose to find $α$ first. To that end, we use the formula $\cos (α) = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$ and substitute $a equals 7$ , $b = \sqrt{53 - 28 \cos (50^{\circ})}$ and $c equals 2$ . We get³
$\cos (α) = \frac{2 - 7 \cos (50^{\circ})}{\sqrt{53 - 28 \cos (50^{\circ})}}$
Since $α$ is an angle in a triangle, we know the radian measure of $α$ must lie between $0$ and $π$ radians. This matches the range of the arccosine function, so we have
$α = \arccos (\frac{2 - 7 \cos (50^{\circ})}{\sqrt{53 - 28 \cos (50^{\circ})}}) radians \approx {114.99}^{\circ}$
At this point, we could find $γ$ using $γ = 180^{\circ} - α - β \approx 180^{\circ} - {114.99}^{\circ} - 50^{\circ} = {15.01}^{\circ}$ , that is if we trust our approximation for $α$ . To minimize propagation of error, however, we could use the Law of Cosines again,⁴ in this case using $\cos (γ) = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$ . Plugging in $a equals 7$ , $b = \sqrt{53 - 28 \cos (50^{\circ})}$ and $c equals 2$ , we get $γ = \arccos (\frac{7 - 2 \cos (50^{\circ})}{\sqrt{53 - 28 \cos (50^{\circ})}})$ radians $approximately equals 15.01 degrees$ . We sketch the triangle below.
Figure 11.49
As we mentioned earlier, once we've determined $b$ it is possible to use the Law of Sines to find the remaining angles. Here, however, we must proceed with caution as we are in the ambiguous (ASS) case. It is advisable to first find the smallest of the unknown angles, since we are guaranteed it will be acute.⁵ In this case, we would find $γ$ since the side opposite $γ$ is smaller than the side opposite the other unknown angle, $α$ . Using the angle-side opposite pair $(β, b)$ , we get $\frac{\sin (γ)}{2} = \frac{\sin (50^{\circ})}{\sqrt{53 - 28 \cos (50^{\circ})}}$ . The usual calculations produces $γ approximately equals 15.01 degrees$ and $α = 180^{\circ} - β - γ \approx 180^{\circ} - 50^{\circ} - {15.01}^{\circ} = {114.99}^{\circ}$ .
Since all three sides and no angles are given, we are forced to use the Law of Cosines. Following our discussion in the previous problem, we find $β$ first, since it is opposite the longest side, $b$ . We get $\cos (β) = \frac{a^{2} + c^{2} - b^{2}}{2 a c} = - \frac{1}{5}$ , so we get $β = \arccos (- \frac{1}{5})$ radians $approximately equals 101.54 degrees$ . As in the previous problem, now that we have obtained an angle-side opposite pair $(β, b)$ , we could proceed using the Law of Sines. The Law of Cosines, however, offers us a rare opportunity to find the remaining angles using only the data given to us in the statement of the problem. Using this, we get $γ = \arccos (\frac{5}{7})$ radians $approximately equals 44.42 degrees$ and $α = \arccos (\frac{29}{35})$ radians $approximately equals 34.05 degrees$ .
Figure 11.50

We note that, depending on how many decimal places are carried through successive calculations, and depending on which approach is used to solve the problem, the approximate answers you obtain may differ slightly from those the authors obtain in the Examples and the Exercises. A great example of this is number in Example Example 1, where the approximate values we record for the measures of the angles sum to $180.01 degrees$ , which is geometrically impossible. Next, we have an application of the Law of Cosines.

Example 2

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. If the angle between the two lines of sight is $60 degrees$ , find the width of the pond.

Figure 11.51

We are given the lengths of two sides and the measure of an included angle, so we may apply the Law of Cosines to find the length of the missing side opposite the given angle. Calling this length $w$ (for width), we get $w^{2} = 950^{2} + 1000^{2} - 2 (950) (1000) \cos (60^{\circ}) = 952500$ from which we get $w = \sqrt{952500} \approx 976$ feet.

In Section, we used the proof of the Law of Sines to develop Theorem as an alternate formula for the area enclosed by a triangle. In this section, we use the Law of Cosines to derive another such formula - Heron's Formula.

Theorem

Heron's Formula: Suppose $a$ , $b$ and $c$ denote the lengths of the three sides of a triangle. Let $s$ be the semiperimeter of the triangle, that is, let $s = \frac{1}{2} (a + b + c)$ . Then the area $A$ enclosed by the triangle is given by

$A = \sqrt{s (s - a) (s - b) (s - c)}$

We prove Theorem using Theorem. Using the convention that the angle $γ$ is opposite the side $c$ , we have $A = \frac{1}{2} a b \sin (γ)$ from Theorem. In order to simplify computations, we start by manipulating the expression for $A^{2}$ .

$\begin{matrix} A^{2} & = & {(\frac{1}{2} a b \sin (γ))}^{2} \\ = & \frac{1}{4} a^{2} b^{2} \sin^{2} (γ) \\ = & \frac{a^{2} b^{2}}{4} (1 - \cos^{2} (γ)) & since \sin^{2} (γ) = 1 - \cos^{2} (γ) . \end{matrix}$

The Law of Cosines tells us $\cos (γ) = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$ , so substituting this into our equation for $A^{2}$ gives

$\begin{matrix} A^{2} & = & \frac{a^{2} b^{2}}{4} (1 - \cos^{2} (γ)) & perfect square trinomials. \\ = & \frac{a^{2} b^{2}}{4} [1 - {(\frac{a^{2} + b^{2} - c^{2}}{2 a b})}^{2}] \\ = & \frac{a^{2} b^{2}}{4} [1 - \frac{{(a^{2} + b^{2} - c^{2})}^{2}}{4 a^{2} b^{2}}] \\ = & \frac{a^{2} b^{2}}{4} [\frac{4 a^{2} b^{2} - {(a^{2} + b^{2} - c^{2})}^{2}}{4 a^{2} b^{2}}] \\ = & \frac{4 a^{2} b^{2} - {(a^{2} + b^{2} - c^{2})}^{2}}{16} \\ = & \frac{{(2 a b)}^{2} - {(a^{2} + b^{2} - c^{2})}^{2}}{16} \\ = & \frac{(2 a b - [a^{2} + b^{2} - c^{2}]) (2 a b + [a^{2} + b^{2} - c^{2}])}{16} & difference of squares. \\ = & \frac{(c^{2} - a^{2} + 2 a b - b^{2}) (a^{2} + 2 a b + b^{2} - c^{2})}{16} \end{matrix}$

$\begin{matrix} A^{2} & = & \frac{(c^{2} - [a^{2} - 2 a b + b^{2}]) ([a^{2} + 2 a b + b^{2}] - c^{2})}{16} \\ = & \frac{(c^{2} - {(a - b)}^{2}) ({(a + b)}^{2} - c^{2})}{16} & perfect square trinomials. \\ = & \frac{(c - (a - b)) (c + (a - b)) ((a + b) - c) ((a + b) + c)}{16} & difference of squares. \\ = & \frac{(b + c - a) (a + c - b) (a + b - c) (a + b + c)}{16} \\ = & \frac{(b + c - a)}{2} \cdot \frac{(a + c - b)}{2} \cdot \frac{(a + b - c)}{2} \cdot \frac{(a + b + c)}{2} \end{matrix}$

At this stage, we recognize the last factor as the semiperimeter, $s = \frac{1}{2} (a + b + c) = \frac{a + b + c}{2}$ . To complete the proof, we note that

$(s - a) = \frac{a + b + c}{2} - a = \frac{a + b + c - 2 a}{2} = \frac{b + c - a}{2}$

Similarly, we find $(s - b) = \frac{a + c - b}{2}$ and $(s - c) = \frac{a + b - c}{2}$ . Hence, we get

$\begin{matrix} A^{2} & = & \frac{(b + c - a)}{2} \cdot \frac{(a + c - b)}{2} \cdot \frac{(a + b - c)}{2} \cdot \frac{(a + b + c)}{2} \\ = & (s - a) (s - b) (s - c) s \end{matrix}$

so that $A = \sqrt{s (s - a) (s - b) (s - c)}$ as required.

We close with an example of Heron's Formula.

Example 3

Find the area enclosed of the triangle in Example Example 1 number.

We are given $a equals 4$ , $b equals 7$ and $c equals 5$ . Using these values, we find $s = \frac{1}{2} (4 + 7 + 5) = 8$ , $(s - a) = 8 - 4 = 4$ , $(s - b) = 8 - 7 = 1$ and $(s - c) = 8 - 5 = 3$ . Using Heron's Formula, we get $A = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{(8) (4) (1) (3)} = \sqrt{96} = 4 \sqrt{6} \approx 9.80$ square units.

Exercises

In Exercises -, use the Law of Cosines to find the remaining side(s) and angle(s) if possible.

$a = 7, b = 12, γ = {59.3}^{\circ}$
$α = 104^{\circ}, b = 25, c = 37$
$a = 153, β = {8.2}^{\circ}, c = 153$
$a = 3, b = 4, γ = 90^{\circ}$
$α = 120^{\circ}, b = 3, c = 4$
$a = 7, b = 10, c = 13$
$a = 1, b = 2, c = 5$
$a = 300, b = 302, c = 48$
$a = 5, b = 5, c = 5$
$a = 5, b = 12,; c = 13$
$a = 18, α = 63^{\circ}, b = 20$
$a = 37, b = 45, c = 26$
$a = 16, α = 63^{\circ}, b = 20$
$a = 22, α = 63^{\circ}, b = 20$
$α = 42^{\circ}, b = 117, c = 88$
$β = 7^{\circ}, γ = 170^{\circ}, c = 98.6$
Find the area of the triangles given in Exercises, and above.
The hour hand on my antique Seth Thomas schoolhouse clock in 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.
A geologist wants to measure the diameter of a crater. From her camp, it is 4 miles to the northern-most point of the crater and 2 miles to the southern-most point. If the angle between the two lines of sight is $117 degrees$ , what is the diameter of the crater? Round your answer to the nearest hundredth of a mile.
From the Pedimaxus International Airport a tour helicopter can fly to Cliffs of Insanity Point by following a bearing of N $8.2 degrees$ E for 192 miles and it can fly to Bigfoot Falls by following a bearing of S $68.5 degrees$ E for 207 miles.⁶ Find the distance between Cliffs of Insanity Point and Bigfoot Falls. Round your answer to the nearest mile.
Cliffs of Insanity Point and Bigfoot Falls from Exericse above both lie on a straight stretch of the Great Sasquatch Canyon. What bearing would the tour helicopter need to follow to go directly from Bigfoot Falls to Cliffs of Insanity Point? Round your angle to the nearest tenth of a degree.
A naturalist sets off on a hike from a lodge on a bearing of S $80 degrees$ W. After 1.5 miles, she changes her bearing to S $17 degrees$ W and continues hiking for 3 miles. Find her distance from the lodge at this point. Round your answer to the nearest hundredth of a mile. What bearing should she follow to return to the lodge? Round your angle to the nearest degree.
The HMS Sasquatch leaves port on a bearing of N $23 degrees$ E and travels for 5 miles. It then changes course and follows a heading of S $41 degrees$ E for 2 miles. How far is it from port? Round your answer to the nearest hundredth of a mile. What is its bearing to port? Round your angle to the nearest degree.
The SS Bigfoot leaves a harbor bound for Nessie Island which is 300 miles away at a bearing of N $32 degrees$ E. A storm moves in and after 100 miles, the captain of the Bigfoot finds he has drifted off course. If his bearing to the harbor is now S $70 degrees$ W, how far is the SS Bigfoot from Nessie Island? Round your answer to the nearest hundredth of a mile. What course should the captain set to head to the island? Round your angle to the nearest tenth of a degree.
From a point 300 feet above level ground in a firetower, a ranger spots two fires in the Yeti National Forest. The angle of depression⁷ made by the line of sight from the ranger to the first fire is $2.5 degrees$ and the angle of depression made by line of sight from the ranger to the second fire is $1.3 degrees$ . The angle formed by the two lines of sight is $117 degrees$ . Find the distance between the two fires. Round your answer to the nearest foot. (Hint: In order to use the $117 degrees$ angle between the lines of sight, you will first need to use right angle Trigonometry to find the lengths of the lines of sight. This will give you a Side-Angle-Side case in which to apply the Law of Cosines.)
Figure 11.52
If you apply the Law of Cosines to the ambiguous Angle-Side-Side (ASS) case, the result is a quadratic equation whose variable is that of the missing side. If the equation has no positive real zeros then the information given does not yield a triangle. If the equation has only one positive real zero then exactly one triangle is formed and if the equation has two distinct positive real zeros then two distinct triangles are formed. Apply the Law of Cosines to Exercises, and above in order to demonstrate this result.
Discuss with your classmates why Heron's Formula yields an area in square units even though four lengths are being multiplied together.

In Exercises -, solve for the remaining side(s) and angle(s), if possible, using any appropriate technique.

Answers

$\begin{matrix} α \approx {35.54}^{\circ} & β \approx {85.16}^{\circ} & γ = {59.3}^{\circ} \\ a = 7 & b = 12 & c \approx 10.36 \end{matrix}$
$\begin{matrix} α = 104^{\circ} & β \approx {29.40}^{\circ} & γ \approx {46.60}^{\circ} \\ a \approx 49.41 & b = 25 & c = 37 \end{matrix}$
$\begin{matrix} α \approx {85.90}^{\circ} & β = {8.2}^{\circ} & γ \approx {85.90}^{\circ} \\ a = 153 & b \approx 21.88 & c = 153 \end{matrix}$
$\begin{matrix} α \approx {36.87}^{\circ} & β \approx {53.13}^{\circ} & γ = 90^{\circ} \\ a = 3 & b = 4 & c = 5 \end{matrix}$
$\begin{matrix} α = 120^{\circ} & β \approx {25.28}^{\circ} & γ \approx {34.72}^{\circ} \\ a = \sqrt{37} & b = 3 & c = 4 \end{matrix}$
$\begin{matrix} α \approx {32.31}^{\circ} & β \approx {49.58}^{\circ} & γ \approx {98.21}^{\circ} \\ a = 7 & b = 10 & c = 13 \end{matrix}$
Table 11.9
Information does not
produce a triangle
$\begin{matrix} α \approx {83.05}^{\circ} & β \approx {87.81}^{\circ} & γ \approx {9.14}^{\circ} \\ a = 300 & b = 302 & c = 48 \end{matrix}$
$\begin{matrix} α = 60^{\circ} & β = 60^{\circ} & γ = 60^{\circ} \\ a = 5 & b = 5 & c = 5 \end{matrix}$
$\begin{matrix} α \approx {22.62}^{\circ} & β \approx {67.38}^{\circ} & γ = 90^{\circ} \\ a = 5 & b = 12 & c = 13 \end{matrix}$
$\begin{matrix} α = 63^{\circ} & β \approx {98.11}^{\circ} & γ \approx {18.89}^{\circ} \\ a = 18 & b = 20 & c \approx 6.54 \end{matrix}$
$\begin{matrix} α = 63^{\circ} & β \approx {81.89}^{\circ} & γ \approx {35.11}^{\circ} \\ a = 18 & b = 20 & c \approx 11.62 \end{matrix}$
$\begin{matrix} α \approx {55.30}^{\circ} & β \approx {89.40}^{\circ} & γ \approx {35.30}^{\circ} \\ a = 37 & b = 45 & c = 26 \end{matrix}$
Table 11.10
Information does not
produce a triangle
$\begin{matrix} α = 63^{\circ} & β \approx {54.1}^{\circ} & γ \approx {62.9}^{\circ} \\ a = 22 & b = 20 & c \approx 21.98 \end{matrix}$
$\begin{matrix} α = 42^{\circ} & β \approx {89.23}^{\circ} & γ \approx {48.77}^{\circ} \\ a \approx 78.30 & b = 117 & c = 88 \end{matrix}$
$\begin{matrix} α \approx 3^{\circ} & β = 7^{\circ} & γ = 170^{\circ} \\ a \approx 29.72 & b \approx 69.2 & c = 98.6 \end{matrix}$
The area of the triangle given in Exercise is $\sqrt{1200} = 20 \sqrt{3} \approx 34.64$ square units. The area of the triangle given in Exercise is $\sqrt{51764375} \approx 7194.75$ square units. The area of the triangle given in Exercise is exactly $30$ square units.
The distance between the ends of the hands at four o'clock is about $8.26$ inches.
The diameter of the crater is about 5.22 miles.
About 313 miles
N $31.8 degrees$ W
She is about 3.92 miles from the lodge and her bearing to the lodge is N $37 degrees$ E.
It is about 4.50 miles from port and its heading to port is S $47 degrees$ W.
It is about 229.61 miles from the island and the captain should set a course of N $16.4 degrees$ E to reach the island.
The fires are about 17456 feet apart. (Try to avoid rounding errors.)

Adapted from Precalculus, 3rd corrected edition, by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.