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The Law of Cosines

The law of cosines can be used to solve for angles and sides of a triangle in cases where other laws do not apply.

Learning Objective

  • Use the law of cosines to solve problems with triangles of any configuration, as well as to transform trigonometric expressions


Key Points

    • The law of cosines can be used to find the measurements of angles and sides of a triangle in cases where the law of sines cannot be applied, such as for triangles with no known angles.
    • 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 example, $a^2 = b^2 + c^2 - 2bc\cos\alpha $).
    • The law of cosines has noted similarities to the Pythagorean theorem, from which it is derived.

Term

  • Pythagorean theorem

    A fundamental relation among the sides of a right triangle, which states that $a^2 + b^2 = c^2$ where $c$ is the hypotenuse, and $a$ and $b$ are the lengths of the other two sides.


Full Text

Using The Law of Cosines

In some cases, we may not have enough information to apply the Law of Sines to find unknown angles and sides in a triangle. For example, consider a triangle where all three sides are known, but no angle values are known. In such cases, there is not enough information to use the Law of Sines. The Law of Cosines is useful for: 1) computing the third side of a triangle when two sides and their enclosed angle are known, and 2) computing the angles of a triangle if only the three sides are known.

The Law of Cosines 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 many formulas at this mathematical level.

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 with angles $\alpha$, $\beta$, and $\gamma$, and opposite corresponding sides $a$, $b$, and $c$, respectively, the Law of Cosines is given as three equations:

$\begin{aligned} a^2 &= b^2 + c^2 - 2bc\cos\alpha \\ b^2 &= a^2 + c^2 - 2ac\cos\beta \\ c^2 &= a^2 + b^2 - 2ab\cos\gamma \end{aligned}$

Oblique triangle

An oblique triangle, with angles $\alpha$, $\beta$, and $\gamma$, and opposite corresponding sides $a$, $b$, and $c$.

To solve for a missing side measurement, the corresponding opposite angle measure is needed. When solving for an angle, the lengths of all of the sides are needed. Notice that each formula for the Law of Cosines can be rearranged to solve for the angle. For example, to solve for the angle $\alpha$, the first formula can be rewritten as:

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

The Law of Cosines is a more general form of the Pythagorean theorem, which holds only for right triangles. Notice that if any angle $\theta$ in the triangle is a right angle (of measure $90^{\circ}$), then $\cos \theta = 0$, and the last term in the Law of Cosines cancels. Thus, for right triangles, the Law of Cosines reduces to the Pythagorean theorem:

 $c^{2}=a^{2}+b^{2}$

Example

Find the length of the unknown side $b$ of the triangle in the figure shown, and round the value to the nearest tenth.

Example oblique triangle

This oblique triangle has known side lengths $a=10$ and $c=12$, and known angle $\beta = 30^{\circ}$.

First, make note of what is given: two sides and the angle between them. This is not enough information to solve the problem using the Law of Sines, but we have the information needed to apply the Law of Cosines.  

We can use the Law of Cosines to solve for side $b$, since we know the measurement of the opposite angle $\beta$. The appropriate formula is:  

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

Substitute the values of $a$, $c$, and $\beta$ from the given triangle:

$b^2 = 10^2 + 12^2 - 2(10)(12)\cos{\left(30^{\circ}\right) }$

From the unit circle, we find that $\displaystyle{\cos{\left(30^{\circ}\right)} = \frac{\sqrt{3}}{2}}$. Substitute this into the formula and evaluate:

$\displaystyle{ \begin{aligned} b^2 &= 10^2 + 12^2 - 2\left(10\right)\left(12\right)\left(\frac{\sqrt{3}}{2}\right) \\ b^2 &= 100 + 144 - 240\frac{\sqrt{3}}{2} \\ b^2 &= 244 - 120\sqrt{3} \\ b &= \sqrt{244 - 120\sqrt{3}} \\ b &\approx 6.0 \end{aligned} }$

Note that we now have enough information that we could use the Law of Sines to solve for the unknown angles $\alpha$ and $\gamma$ in the triangle. 

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