Law of Cosines & Heron's Formula Lesson

Over the course of the last two lessons, we have learned how to use the law of sines to solve oblique triangles given ASA, SAA, and SSA. Now, we will learn how to solve oblique triangles given two sides and the included angle (SAS), or three sides of the triangle (SSS). When both of these cases are presented, a unique triangle is determined, which can be solved using the law of cosines.

Derivation of the Law of Cosines

First, let's think about the triangle side length restriction. In any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side.
As an example, it would be impossible to construct a triangle with sides of lengths 8, 9, and 20. $$8 + 9 < 20$$ To derive the law of cosines, let ABC be any oblique triangle. We will place the vertex B at the origin of our coordinate plane and side BC along the positive x-axis. If we let (x, y) be the coordinates of the vertex A of our triangle, then the following will be true for any angle B, whether obtuse or acute: $$\text{sin}\hspace{.1em}B=\frac{y}{c}$$ If we solve for y: $$y=c \cdot \text{sin}\hspace{.1em}B$$ $$\text{cos}\hspace{.1em}B=\frac{x}{c}$$ If we solve for x: $$x=c \cdot \text{cos}\hspace{.1em}B$$ As you can see from our image above, the coordinates of the point are labeled as: $$(c \cdot \text{cos}\hspace{.1em}B, c \cdot \text{sin}\hspace{.1em}B)$$ Additionally, we can see that point C has the coordinates: $$(a, 0)$$ Side AC has length b (since it is opposite angle B).
Recall the distance formula, which is derived from the Pythagorean theorem: $$d=\sqrt{(x_{2}- x_{1})^2 + (y_{2}- y_{1})^2}$$ We can use the distance formula here, to write an equation: If we use the following to plug into the distance formula: $$x_{1}=a$$ $$x_{2}=c \cdot \text{cos}\hspace{.1em}B$$ $$y_{1}=0$$ $$y_{2}=c \cdot \text{sin}\hspace{.1em}B$$ $$b=\sqrt{(c \cdot \text{cos}\hspace{.1em}B - a)^2 + (c \cdot \text{sin}\hspace{.1em}B - 0)^2}$$ Let's now square both sides and simplify: $$b^2=(c \cdot \text{cos}\hspace{.1em}B - a)^2 + (c \cdot \text{sin}\hspace{.1em}B)^2$$ $$b^2=c^2 \hspace{.1em}\text{cos}^{2}\hspace{.1em}B - 2ac \hspace{.1em}\text{cos}\hspace{.1em}B + a^2 + c^2 \hspace{.1em}\text{sin}^{2}\hspace{.1em}B$$ In order to take advantage of one of our Pythagorean identities, let's rearrange things: $$b^2=a^2 + c^2 \hspace{.1em}\text{cos}^{2}\hspace{.1em}B + c^2 \hspace{.1em}\text{sin}^{2}\hspace{.1em}B - 2ac \hspace{.1em}\text{cos}\hspace{.1em}B$$ Let's factor out the c² from the middle two terms: $$b^2=a^2 + c^2(\text{cos}^{2}\hspace{.1em}B + \text{sin}^{2}\hspace{.1em}B) - 2ac \hspace{.1em}\text{cos}\hspace{.1em}B$$ Recall the Pythagorean Identity: $$\text{sin}^{2}θ + \text{cos}^2 θ=1$$ Let's use this identity: $$(\text{cos}^{2}\hspace{.1em}B + \text{sin}^{2}\hspace{.1em}B)=1$$ Replace this in our equation: $$b^2=a^2 + c^2(1) - 2ac \hspace{.1em}\text{cos}\hspace{.1em}B$$ Simplify: $$b^2=a^2 + c^2 - 2ac \hspace{.1em}\text{cos}\hspace{.1em}B$$ The result is one form of the law of cosines. We can use a similar process to derive the other forms:

Law of Cosines

In any triangle ABC, with sides a, b, and c: $$a^2=b^2 + c^2 - 2bc \hspace{.1em}\text{cos}\hspace{.1em}A$$ $$b^2=a^2 + c^2 - 2ac \hspace{.1em}\text{cos}\hspace{.1em}B$$ $$c^2=a^2 + b^2 - 2ab \hspace{.1em}\text{cos}\hspace{.1em}C$$ To summarize our law of cosines, we can say that the square of a side of a triangle is equal to the sum of the squares of the two other sides less twice the product of the two sides and the cosine of the included angle. Let's begin by looking at a few examples where we are given SAS.

Law of Cosines (SAS) Examples

1. Find the missing side using the law of cosines
2. Find the smaller of the two remaining angles using the law of sines
   • A triangle can only have one obtuse angle
   • An obtuse angle in a triangle must be opposite the largest side
   • The inverse sine function can't give us obtuse angles
3. Find the remaining angle using the angle sum property of a triangle

Example #1: Solve each triangle. Round your answer to the nearest tenth.
a = 27 in, b = 21 in, and C = 100°.
Let's create a simple sketch. Let's grab the appropriate formula from above: $$c^2=a^2 + b^2 - 2ab \hspace{.1em}\text{cos}\hspace{.1em}C$$ Let's plug into our formula: $$c^2=(27 \hspace{.1em}\text{in})^2 + (21 \hspace{.1em}\text{in})^2 - 2(27 \hspace{.1em}\text{in})(21 \hspace{.1em}\text{in}) \hspace{.1em}\text{cos}\hspace{.1em}100°$$ $$c^2=729\hspace{.1em}\text{in}^{2}+ 441\hspace{.1em}\text{in}^{2}- 1134\hspace{.1em}\text{in}^{2}\hspace{.1em}\text{cos}\hspace{.1em}100°$$ $$c^2=1170\hspace{.1em}\text{in}^{2}- 1134\hspace{.1em}\text{in}^{2}\hspace{.1em}\text{cos}\hspace{.1em}100°$$ Since we want to solve for c, let's take the square root of each side. Note, we only need the principal square root on the right: $$c=\sqrt{1170 \hspace{.1em}\text{in}^{2}- 1134\text{in}^{2}\hspace{.1em}\text{cos}\hspace{.1em}100°}$$ $$c ≈ 37 \hspace{.1em}\text{in}$$ Now that we have our missing side, we can use the law of sines to find the smaller of the two missing angles, then use the angle sum property of a triangle to find the other. $$\frac{\text{sin}\hspace{.1em}C}{c}=\frac{\text{sin}\hspace{.1em}B}{b}$$ Let's plug in for C, c, and b. Note, c is an approximation. $$\frac{\text{sin}\hspace{.1em}100°}{37 \hspace{.1em}\text{in}}=\frac{\text{sin}\hspace{.1em}B}{21 \hspace{.1em}\text{in}}$$ $$\text{sin}\hspace{.1em}B=21 \hspace{.1em}\text{in}\cdot \frac{\text{sin}\hspace{.1em}100°}{37 \hspace{.1em}\text{in}}$$ $$\text{sin}\hspace{.1em}B ≈ .5589$$ Let's use the inverse sine function to find B: $$B ≈ \text{sin}^{-1}(.5589) ≈ 34°$$ To find A, let's subtract 180° - C - B: $$180° - 100° - 34°=46°$$ $$A ≈ 46°$$ Example #2: Solve each triangle. Round your answer to the nearest tenth.
a = 28 cm, c = 16 cm, and B = 89°.
Let's create a simple sketch. Let's grab the appropriate formula from above: $$b^2=a^2 + c^2 - 2ac \hspace{.1em}\text{cos}\hspace{.1em}B$$ Let's plug into our formula: $$c^2=(28 \hspace{.1em}\text{cm})^2 + (16 \hspace{.1em}\text{cm})^2 - 2(28 \hspace{.1em}\text{cm})(16 \hspace{.1em}\text{cm}) \hspace{.1em}\text{cos}\hspace{.1em}89°$$ $$b^2=784\hspace{.1em}\text{cm}^{2}+ 256\hspace{.1em}\text{cm}^{2}- 896\hspace{.1em}\text{cm}^{2}\hspace{.1em}\text{cos}\hspace{.1em}89°$$ $$b^2=1040 \hspace{.1em}\text{cm}^{2}- 896\text{cm}^{2}\hspace{.1em}\text{cos}\hspace{.1em}89°$$ Since we want to solve for b, let's take the square root of each side. Note, we only need the principal square root on the right: $$b=\sqrt{1040 \hspace{.1em}\text{cm}^{2}- 896\text{cm}^{2}\hspace{.1em}\text{cos}\hspace{.1em}89°}$$ $$b ≈ 32 \hspace{.1em}\text{cm}$$ Now that we have our missing side, we can use the law of sines to find the smaller of the two missing angles, then use the angle sum property of a triangle to find the other. $$\frac{\text{sin}\hspace{.1em}B}{b}=\frac{\text{sin}\hspace{.1em}C}{c}$$ Let's plug in for B, b, and c. Note, b is an approximation. $$\frac{\text{sin}\hspace{.1em}89°}{32 \hspace{.1em}\text{cm}}=\frac{\text{sin}\hspace{.1em}C}{16 \hspace{.1em}\text{cm}}$$ $$\text{sin}\hspace{.1em}C=16 \hspace{.1em}\text{cm}\cdot \frac{\text{sin}\hspace{.1em}89°}{32 \hspace{.1em}\text{cm}}$$ $$\text{sin}\hspace{.1em}C ≈ .4999$$ Let's use the inverse sine function to find C: $$C ≈ \text{sin}^{-1}(.4999) ≈ 30°$$ To find A, let's subtract 180° - B - C: $$180° - 89° - 30°=61°$$ $$A ≈ 61°$$ Let's now look at a few examples where we are given all three sides.

Law of Cosines (SSS) Examples

1. Find the largest angle using the law of cosines
2. Find either remaining angle using the law of sines
3. Find the remaining angle using the angle sum property of a triangle

Example #3: Solve each triangle. Round your answer to the nearest tenth.
a = 17 mi, b = 15 mi, c = 28 mi
Let's create a simple sketch. We can use the law of cosines to solve for any angle of a triangle. Since c is the largest side, it is opposite of C, the largest angle. We know that C is obtuse if cos C < 0 (recall in quadrant II, cosine values are negative). Let's start with our formula that involves the cosine of C: $$c^2=a^2 + b^2 - 2ab \hspace{.1em}\text{cos}\hspace{.1em}C$$ Solve for cos C: $$c^2 - a^2 - b^2=- 2ab \hspace{.1em}\text{cos}\hspace{.1em}C$$ $$\frac{c^2 - a^2 - b^2}{-2ab}=\frac{- 2ab \hspace{.1em}\text{cos}\hspace{.1em}C}{-2ab}$$ $$\text{cos}\hspace{.1em}C=\frac{c^2 - a^2 - b^2}{-2ab}$$ $$\text{cos}\hspace{.1em}C=\frac{-c^2 + a^2 + b^2}{2ab}$$ $$\text{cos}\hspace{.1em}C=\frac{a^2 + b^2 - c^2}{2ab}$$ $$\text{cos}\hspace{.1em}C=\frac{(17 \hspace{.1em}\text{mi})^{2}+ (15 \hspace{.1em}\text{mi})^{2}- (28 \hspace{.1em}\text{mi})^{2}}{2(17 \hspace{.1em}\text{mi})(15 \hspace{.1em}\text{mi})}$$ $$\text{cos}\hspace{.1em}C=\frac{289 \hspace{.1em}\text{mi}^{2}+ 225 \hspace{.1em}\text{mi}^{2}- 784 \hspace{.1em}\text{mi}^{2}}{510 \text{mi}^{2}}$$ $$\text{cos}\hspace{.1em}C=\frac{-270 \hspace{.1em}\text{mi}^{2}}{510 \text{mi}^{2}}$$ $$\text{cos}\hspace{.1em}C=-\frac{270 \hspace{.1em}}{510}$$ $$\text{cos}\hspace{.1em}C ≈ -.5294$$ Let's use our inverse cosine function: $$C ≈ \text{cos}^{-1}(-.5294) ≈ 122°$$ Now, let's use the law of sines to find one of our other angles here. $$\frac{\text{sin}\hspace{.1em}C}{c}=\frac{\text{sin}\hspace{.1em}A}{a}$$ $$\frac{\text{sin}\hspace{.1em}122°}{28 \hspace{.1em}\text{mi}}=\frac{\text{sin}\hspace{.1em}A}{17 \hspace{.1em}\text{mi}}$$ $$\text{sin}\hspace{.1em}A=17 \hspace{.1em}\text{mi}\cdot \frac{\text{sin}\hspace{.1em}122°}{28 \hspace{.1em}\text{mi}}$$ $$\text{sin}\hspace{.1em}A ≈ .5149$$ Let's use our inverse sine function: $$A ≈ \text{sin}^{-1}(.5149) ≈ 31°$$ $$180° - 122° - 31°=27°$$ $$B ≈ 27°$$ Example #4: Solve each triangle. Round your answer to the nearest tenth.
a = 13 ft, b = 29 ft, c = 21 ft
Let's create a simple sketch. We can use the law of cosines to solve for any angle of a triangle. Since b is the largest side, it is opposite of B, the largest angle. We know that B is obtuse if cos B < 0 (recall in quadrant II, cosine values are negative). Let's start with our formula that involves the cosine of B: $$b^2=a^2 + c^2 - 2ac \hspace{.1em}\text{cos}\hspace{.1em}B$$ Solve for cos B: $$\text{cos}\hspace{.1em}B=\frac{a^2 + c^2 - b^2}{2ac}$$ Plug in for a, b, and c: $$\text{cos}\hspace{.1em}B=\frac{(13 \hspace{.1em}\text{ft})^2 + (21 \hspace{.1em}\text{ft})^2 - (29 \hspace{.1em}\text{ft})^2}{2(13 \hspace{.1em}\text{ft})(21 \hspace{.1em}\text{ft})}$$ $$\text{cos}\hspace{.1em}B=\frac{169 \hspace{.1em}\text{ft}^{2}+ 441 \hspace{.1em}\text{ft}^{2}- 841 \hspace{.1em}\text{ft}^{2}}{546 \hspace{.1em}\text{ft}^{2}}$$ $$\text{cos}\hspace{.1em}B=\frac{-231 \hspace{.1em}\text{ft}^{2}}{546 \hspace{.1em}\text{ft}^{2}}$$ $$\text{cos}\hspace{.1em}B=-\frac{231}{546}$$ $$\text{cos}\hspace{.1em}B ≈ -.4231$$ Let's use our inverse cosine function: $$B ≈ \text{cos}^{-1}(-.4231) ≈ 115°$$ Now, let's use the law of sines to find one of our other angles here. $$\frac{\text{sin}\hspace{.1em}B}{b}=\frac{\text{sin}\hspace{.1em}A}{a}$$ $$\frac{\text{sin}\hspace{.1em}115°}{29 \hspace{.1em}\text{ft}}=\frac{\text{sin}\hspace{.1em}A}{13 \hspace{.1em}\text{ft}}$$ $$\text{sin}\hspace{.1em}A=13 \hspace{.1em}\text{ft}\cdot \frac{\text{sin}\hspace{.1em}115°}{29 \hspace{.1em}\text{ft}}$$ $$\text{sin}\hspace{.1em}A ≈ .4063$$ Let's use our inverse sine function: $$A ≈ \text{sin}^{-1}(.4063) ≈ 24°$$ $$180° - 115° - 24°=41°$$ $$C ≈ 41°$$

How to Find the Area of a Triangle Using Heron's Formula

We will derive this formula in the practice test section. A full step-by-step is provided in the practice test solution video.
Heron's Area Formula (SSS):
If a triangle has sides of lengths a, b, and c, with semiperimeter s: $$s=\frac{1}{2}(a + b + c)$$ The area of a triangle can then be found as: $$\text{Area}=\sqrt{s(s - a)(s - b)(s - c)}$$ Let's look at a few examples.
Example #5: Find the area of each triangle. Round your answer to the nearest tenth.
a = 10.7 m, b = 9 m, c = 10.5 m.
Let's create a simple sketch. Let's find the semiperimeter s: $$s=\frac{1}{2}(10.7 \hspace{.1em}\text{m}+ 9 \hspace{.1em}\text{m}+ 10.5 \hspace{.1em}\text{m})$$ $$s=\frac{1}{2}(30.2 \hspace{.1em}\text{m})$$ $$s=15.1 \hspace{.1em}\text{m}$$ Let's now plug into Heron's formula: $$\text{Area}=\sqrt{15.1 \hspace{.1em}\text{m}(15.1 \hspace{.1em}\text{m}- 10.7 \hspace{.1em}\text{m})(15.1 \hspace{.1em}\text{m}- 9 \hspace{.1em}\text{m})(15.1 \hspace{.1em}\text{m}- 10.5 \hspace{.1em}\text{m})}$$ Simplify: $$\text{Area}=\sqrt{15.1 \hspace{.1em}\text{m}(4.4 \hspace{.1em}\text{m})(6.1 \hspace{.1em}\text{m})(4.6 \hspace{.1em}\text{m})}$$ $$\text{Area}=\sqrt{1864.3064 \hspace{.1em}\text{m}^{4}}$$ $$\text{Area}≈43.2\hspace{.1em}\text{m}^{2}$$ Example #6: Find the area of each triangle. Round your answer to the nearest tenth.
b = 7 m, A = 71°, c = 5 m
Let's create a simple sketch. Here, we are given SAS. When this happens, it is much more efficient to use the area formula from the law of sines section. In order to get some extra practice, let's solve this problem using Heron's formula.
Since we don't have all three sides, we first want to use the law of cosines to find the length of side a: $$a^2=b^2 + c^2 - 2bc \hspace{.1em}\text{cos}\hspace{.1em}A$$ $$a^2=(7 \hspace{.1em}\text{m})^2 + (5 \hspace{.1em}\text{m})^2 - 2(7 \hspace{.1em}\text{m})(5 \hspace{.1em}\text{m}) \hspace{.1em}\text{cos}\hspace{.1em}71°$$ $$a^2=49 \hspace{.1em}\text{m}^{2}+ 25 \hspace{.1em}\text{m}^{2}- 70 \hspace{.1em}\text{m}^{2}\hspace{.1em}\text{cos}\hspace{.1em}71°$$ $$a^2=74 \hspace{.1em}\text{m}^{2}- 70 \hspace{.1em}\text{m}^{2}\hspace{.1em}\text{cos}\hspace{.1em}71°$$ $$a=\sqrt{74 \hspace{.1em}\text{m}^{2}- 70 \hspace{.1em}\text{m}^{2}\hspace{.1em}\text{cos}\hspace{.1em}71°}$$ $$a ≈ 7.2 \hspace{.1em}\text{m}$$ Now, we can use Heron's Formula: a = 7.2 m, b = 7 m, c = 5 m. Let's find the semiperimeter s: Note: the value for a is an approximation found above. $$s=\frac{1}{2}(7.2 \hspace{.1em}\text{m}+ 7 \hspace{.1em}\text{m}+ 5 \hspace{.1em}\text{m})$$ $$s=\frac{1}{2}(19.2 \hspace{.1em}\text{m})$$ $$s=9.6 \hspace{.1em}\text{m}$$ Let's now plug into Heron's formula: $$\text{Area}=\sqrt{9.6 \hspace{.1em}\text{m}(9.6 \hspace{.1em}\text{m}- 7.2 \hspace{.1em}\text{m})(9.6 \hspace{.1em}\text{m}- 7 \hspace{.1em}\text{m})(9.6 \hspace{.1em}\text{m}- 5 \hspace{.1em}\text{m})}$$ Simplify: $$\text{Area}=\sqrt{9.6 \hspace{.1em}\text{m}(2.4 \hspace{.1em}\text{m})(2.6 \hspace{.1em}\text{m})(4.6 \hspace{.1em}\text{m})}$$ $$\text{Area}=\sqrt{275.5584 \hspace{.1em}\text{m}^{4}}$$ $$\text{Area}≈ 16.6\hspace{.1em}\text{m}^{2}$$ Note with our other formula we would find this much quicker: $$\text{Area}=\frac{1}{2}bc \cdot \text{sin}\hspace{.1em}A$$ $$\text{Area}=\frac{35 \hspace{.1em}\text{m}^2 \cdot \text{sin}\hspace{.1em}71°}{2}$$ $$\text{Area}≈ 16.5 \text{m}^{2}$$ Note, the difference is due to rounding.

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