Non-right Triangles: Law of Cosines

Learning Outcomes

• Use the Law of Cosines to solve oblique triangles.
• Solve applied problems using the Law of Cosines.
• Use Heron’s formula to ﬁnd the area of a triangle.

Suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles as shown in Figure 1. How far from port is the boat?

Figure 1

Unfortunately, while the Law of Sines enables us to address many non-right triangle cases, it does not help us with triangles where the known angle is between two known sides, a SAS (side-angle-side) triangle, or when all three sides are known, but no angles are known, a SSS (side-side-side) triangle. In this section, we will investigate another tool for solving oblique triangles described by these last two cases.

Using the Law of Cosines to Solve Oblique Triangles

The tool we need to solve the problem of the boat’s distance from the port is the Law of Cosines, which defines the relationship among angle measurements and side lengths in oblique triangles. Three formulas make up the Law of Cosines. At first glance, the formulas may appear complicated because they include many variables. However, once the pattern is understood, the Law of Cosines is easier to work with than most formulas at this mathematical level.

Understanding how the Law of Cosines is derived will be helpful in using the formulas. The derivation begins with the Generalized Pythagorean Theorem, which is an extension of the Pythagorean Theorem to non-right triangles. Here is how it works: An arbitrary non-right triangle $ABC$ is placed in the coordinate plane with vertex $A$ at the origin, side $c$ drawn along the x-axis, and vertex $C$ located at some point $\left(x,y\right)$ in the plane, as illustrated in Figure 2. Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted.

Figure 2

We can drop a perpendicular from $C$ to the x-axis (this is the altitude or height). Recalling the basic trigonometric identities, we know that

$\cos \theta =\frac{x\text{(adjacent)}}{b\text{(hypotenuse)}}\text{ and }\sin \theta =\frac{y\text{(opposite)}}{b\text{(hypotenuse)}}$

In terms of $\theta ,\text{ }x=b\cos \theta$ and $y=b\sin \theta .\text{ }$ The $\left(x,y\right)$ point located at $C$ has coordinates $\left(b\cos \theta ,b\sin \theta \right)$. Using the side $\left(x-c\right)$ as one leg of a right triangle and $y$ as the second leg, we can find the length of hypotenuse $a$ using the Pythagorean Theorem. Thus,

\begin{align} {a}^{2}&={\left(x-c\right)}^{2}+{y}^{2} \\ &={\left(b\cos \theta -c\right)}^{2}+{\left(b\sin \theta \right)}^{2} && \text{Substitute }\left(b\cos \theta \right)\text{ for}x\text{and }\left(b\sin \theta \right)\text{for }y. \\ &=\left({b}^{2}{\cos }^{2}\theta -2bc\cos \theta +{c}^{2}\right)+{b}^{2}{\sin }^{2}\theta && \text{Expand the perfect square}. \\ &={b}^{2}{\cos }^{2}\theta +{b}^{2}{\sin }^{2}\theta +{c}^{2}-2bc\cos \theta && \text{Group terms noting that }{\cos }^{2}\theta +{\sin }^{2}\theta =1. \\ &={b}^{2}\left({\cos }^{2}\theta +{\sin }^{2}\theta \right)+{c}^{2}-2bc\cos \theta && \text{Factor out }{b}^{2}. \\ {a}^{2}&={b}^{2}+{c}^{2}-2bc\cos \theta &&{\cos }^{2}\theta +{\sin }^{2}\theta = 1 \end{align}

The formula derived is one of the three equations of the Law of Cosines. The other equations are found in a similar fashion.

Keep in mind that it is always helpful to sketch the triangle when solving for angles or sides. In a real-world scenario, try to draw a diagram of the situation. As more information emerges, the diagram may have to be altered. Make those alterations to the diagram and, in the end, the problem will be easier to solve.

A General Note: Law of Cosines

The Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle. For triangles labeled as in Figure 3, with angles $\alpha ,\beta$, and $\gamma$, and opposite corresponding sides $a,b$, and $c$, respectively, the Law of Cosines is given as three equations.

${a}^{2}={b}^{2}+{c}^{2}-2bc\cos \alpha$

${b}^{2}={a}^{2}+{c}^{2}-2ac\cos \beta$

${c}^{2}={a}^{2}+{b}^{2}-2ab\cos \gamma$

Figure 3

To solve for a missing side measurement, the corresponding opposite angle measure is needed.

When solving for an angle, the corresponding opposite side measure is needed. We can use another version of the Law of Cosines to solve for an angle.

$\cos \alpha =\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}$

$\cos \beta =\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}$

$\cos\gamma =\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}$

How To: Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle.

1. Sketch the triangle. Identify the measures of the known sides and angles. Use variables to represent the measures of the unknown sides and angles.
2. Apply the Law of Cosines to find the length of the unknown side or angle.
3. Apply the Law of Sines or Cosines to find the measure of a second angle.
4. Compute the measure of the remaining angle.

Example 1: Finding the Unknown Side and Angles of a SAS Triangle

Find the unknown side and angles of the triangle in Figure 4.

Figure 4

Try It

Find the missing side and angles of the given triangle: $\alpha =30^\circ ,b=12,c=24$.

Example 2: Solving for an Angle of a SSS Triangle

Find the angle $\alpha$ for the given triangle if side $a=20$, side $b=25$, and side $c=18$.

Try It

Given $a=5,b=7$, and $c=10$, find the missing angles.

Solving Applied Problems Using the Law of Cosines

Just as the Law of Sines provided the appropriate equations to solve a number of applications, the Law of Cosines is applicable to situations in which the given data fits the cosine models. We may see these in the fields of navigation, surveying, astronomy, and geometry, just to name a few.

Example 3: Using the Law of Cosines to Solve a Communication Problem

On many cell phones with GPS, an approximate location can be given before the GPS signal is received. This is accomplished through a process called triangulation, which works by using the distances from two known points. Suppose there are two cell phone towers within range of a cell phone. The two towers are located 6000 feet apart along a straight highway, running east to west, and the cell phone is north of the highway. Based on the signal delay, it can be determined that the signal is 5050 feet from the first tower and 2420 feet from the second tower. Determine the position of the cell phone north and east of the first tower, and determine how far it is from the highway.

Example 4: Calculating Distance Traveled Using a SAS Triangle

Returning to our problem at the beginning of this section, suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat? The diagram is repeated here in Figure 8.

Figure 8

Using Heron’s Formula to Find the Area of a Triangle

We already learned how to find the area of an oblique triangle when we know two sides and an angle. We also know the formula to find the area of a triangle using the base and the height. When we know the three sides, however, we can use Heron’s formula instead of finding the height. Heron of Alexandria was a geometer who lived during the first century A.D. He discovered a formula for finding the area of oblique triangles when three sides are known.

A General Note: Heron’s Formula

Heron’s formula finds the area of oblique triangles in which sides $a,b$, and $c$ are known.

$\text{Area}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

where $s=\frac{\left(a+b+c\right)}{2}$ is one half of the perimeter of the triangle, sometimes called the semi-perimeter.

Example 5: Using Heron’s Formula to Find the Area of a Given Triangle

Find the area of the triangle in Figure 9 using Heron’s formula.

Figure 9

Try It

Use Heron’s formula to find the area of a triangle with sides of lengths $a=29.7\text{ft},b=42.3\text{ft}$, and $c=38.4\text{ft}$.

Example 6: Applying Heron’s Formula to a Real-World Problem

A Chicago city developer wants to construct a building consisting of artist’s lofts on a triangular lot bordered by Rush Street, Wabash Avenue, and Pearson Street. The frontage along Rush Street is approximately 62.4 meters, along Wabash Avenue it is approximately 43.5 meters, and along Pearson Street it is approximately 34.1 meters. How many square meters are available to the developer? See Figure 10 for a view of the city property.

Figure 10

Try It

Find the area of a triangle given $a=4.38\text{ft},b=3.79\text{ft,}$ and $c=5.22\text{ft}\text{.}$

Key Equations

 Law of Cosines ${a}^{2}={b}^{2}+{c}^{2}-2bc\cos \alpha$ ${b}^{2}={a}^{2}+{c}^{2}-2ac\cos \beta$ ${c}^{2}={a}^{2}+{b}^{2}-2ab\cos \gamma$ Heron’s formula $\text{Area}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ where $s=\frac{\left(a+b+c\right)}{2}$

Key Concepts

• The Law of Cosines defines the relationship among angle measurements and lengths of sides in oblique triangles.
• The Generalized Pythagorean Theorem is the Law of Cosines for two cases of oblique triangles: SAS and SSS. Dropping an imaginary perpendicular splits the oblique triangle into two right triangles or forms one right triangle, which allows sides to be related and measurements to be calculated.
• The Law of Cosines is useful for many types of applied problems. The first step in solving such problems is generally to draw a sketch of the problem presented. If the information given fits one of the three models (the three equations), then apply the Law of Cosines to find a solution.
• Heron’s formula allows the calculation of area in oblique triangles. All three sides must be known to apply Heron’s formula.

Glossary

Law of Cosines
states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle
Generalized Pythagorean Theorem
an extension of the Law of Cosines; relates the sides of an oblique triangle and is used for SAS and SSS triangles