3.3 - The Laws of Sines and Cosines
Contents
Objectives:
- Understand and use the Law of Sines.
- Understand and use the Law of Cosines.
- Recognize potential ambiguities when using the Law of Sines.
- Definitions
- Law of Sines, Law of Cosines
Lesson Guide
Up until now, we have only related trigonometric functions to right triangles. Today, the focus will shift to using our trigonometric toolset to find angles and side lengths in non-right triangles. Included in the lesson plans are derivations for the Law of Sines and the Law of Cosines. Include as many (or as few) derivations as are appropriate for your class: while they should be able to follow all derivations, at the end of the day, it is most important that they know how to use both laws.
Standard Triangle Labeling
Start by introducing the standard triangle labeling to students, shown below; the same letter pair gets assigned for each angle and its opposite side. It is crucial to use this convention when applying the Law of Sines or Cosines.
Derivation of the Law of Sines
Our basic process is to create right triangles from non-right triangles. Sketch a line down from angle $B$, perpendicular to side $b$, obtaining the following:
- From the diagram, we see that $\sin(A)=h/c$, and $\sin(C)=h/a$. Solving for $h$ in both equations, we see that $c\sin(A)=h=a\sin(C)$
- Since $c\sin(A)=a\sin(C)$, we conclude $\frac{\sin(A)}{a}=\frac{\sin(C)}{c}$.
- Note that this is a slightly misleading diagram were $A$ or $C$ an obtuse angle. However, the process will still work (you'll just need to be more careful about reminding students that supplementary angles have the same sine).
- Were we to repeat the steps above, only starting with angle $A$, we would find also that $\frac{\sin(B)}{b}=\frac{\sin(C)}{c}$.
Combining the equations from above, we obtain the Law of Sines:
$\frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{\sin(C)}{c}$.
Have students complete Problems 1 and 2 on their worksheets.
Derivation of the Law of Cosines
Again, sketch a line down from angle $B$, perpendicular to side $b$. Applying the Pythagorean theorem to both of the right triangles this creates, we obtain the following:
- $r^2+h^2=a^2$, and $(b-r)^2+h^2=c^2$.
- Solving these two equations for $h^2$, we obtain $a^2-r^2=h^2=c^2-(b-r)^2$.
- Expanding and simplifying, $a^2+b^2-2br=c^2$.
- Next, by the right-triangle definition of cosine, we also have that $\cos(C)=\frac{r}{a}$, so $a\cos(C)=r$.
Substituting, we obtain the Law of Cosines:
$a^2+b^2-2ab\cos(C) = c^2$
Starting with a different labeling, we may also show each of the following equivalent statements of the Law of Cosines:
$b^2+c^2-2bc\cos(A) = a^2$ $a^2+c^2-2ac\cos(B) = b^2$
Have students complete the worksheet. Problem 5 outlines the potential ambiguity with the Law of Sines. Make sure this is clear to your students once they have had a chance to work through the problem.
Comments
In this area, you should feel free to add any comments you may have on how this lesson has gone or what other instructors should be aware of.
After problem 5, you may want to discuss how we can know if we are in the ambiguous case, and how we can know if there are one or two possible triangles. Namely, the ambiguous case arises when we are given two sides and an angle $\phi$ not between them (SSA). Using Law of Sines to find another angle tells us the other angle is either $\theta$ or $\pi - \theta$. Certainly the angle $\theta$ completes a valid triangle. But if $\pi - \theta + \phi < 180$, then $\pi - \theta$ also gives a valid triangle, so there are two possible triangles.
