The Law of Sines

Return to MAT 130 main page

Go to Preview Activity

---

Textbook Reference: Section 8.1 Non-right Triangles: Law of Sines

---

You now have two important tools to solve right triangles:

• You can use the circular functions (and their inverses) to relate the sides to the angles. (Think SOH-CAH-TOA.)
• You can use the Pythagorean Theorem (\(a^2+b^2=c^2\)) to relate the sides to each other.

But right triangles will only get you so far.

What can we do about oblique triangles — triangles that don't have a right angle anywhere?

It turns out we generalize the above two tools to come up with formulas that work for oblique triangles — a generalization of SOH-CAH-TOA called the Law of Sines, and a generalization of the Pythagorean Theorem called the Law of Cosines.

To keep things consistent, we generally use the following naming conventions:

• Angles will be named with capital letters (or occasionally you'll see Greek letters).
• Sides will be named with lowercase letters.
• Each side and its opposite angle will be named with the same letter.

The Law of Sines

You learned back in high school geometry that larger sides are always across from larger angles. It's natural to wonder whether this is a proportional relationship — does double the angle mean double the side length?

Unfortunately, no. But the Law of Sines is the next best thing: \[\dfrac{\color{flatblue}a}{\sin({\color{flatblue}A})}=\dfrac{\color{flatpurple}b}{\sin({\color{flatpurple}B})}=\dfrac{\color{flatred}c}{\sin({\color{flatred}C})}\]

Even though the above equation involves all three sides, you only ever really need to use two at a time, for whichever two side-angle pairs you want to compare.

If you'd like to see the proof of the Law of Sines, click in the box below to expand it. Otherwise, you're welcome to skip it. (Though you may as well now. You know you want to.)

Proof: Law of Sines

Let's just prove for now that \(\dfrac{\color{flatblue}a}{\sin({\color{flatblue}A})}=\dfrac{\color{flatred}c}{\sin({\color{flatred}C})}\).

Suppose we drop an altitude of length \(h\) from point \(\color{flatpurple}B\) as follows:

We can use SOH-CAH-TOA on the two triangles we have and get two equations: \[\sin({\color{flatblue}A})=\dfrac{h}{\color{flatred}c}\qquad\qquad\sin({\color{flatred}C})=\dfrac{h}{\color{flatblue}a}\] But look! The variable \(h\) appears in both equations, so we could solve for it in both: \[h={\color{flatred}c}\sin({\color{flatblue}A})\qquad\qquad h={\color{flatblue}a}\sin({\color{flatred}C})\] If both expressions equal \(h\), then they must be equal to each other! \[{\color{flatblue}a}\sin({\color{flatred}C}) = {\color{flatred}c}\sin({\color{flatblue}A})\] Finally, divide both sides by the quantity \(\sin({\color{flatblue}A})\sin({\color{flatred}C})\) — think of it like "un-cross-multiplying" — and voilà: \[\dfrac{\color{flatblue}a}{\sin({\color{flatblue}A})}=\dfrac{\color{flatred}c}{\sin({\color{flatred}C})}\] The beautiful thing, though is that there's nothing special about choosing \(\color{flatpurple}B\)! We could have just as easily repeated this argument by turning the triangle on its side and dropping an altitude from \(\color{flatblue}A\) or \(\color{flatred}C\), and we would have gotten \(\dfrac{\color{flatblue}a}{\sin({\color{flatblue}A})}=\dfrac{\color{flatpurple}b}{\sin({\color{flatpurple}B})}\) or \(\dfrac{\color{flatpurple}b}{\sin({\color{flatpurple}B})}=\dfrac{\color{flatred}c}{\sin({\color{flatred}C})}\). So, really, three of these proportions are equal: \[\dfrac{\color{flatblue}a}{\sin({\color{flatblue}A})}=\dfrac{\color{flatpurple}b}{\sin({\color{flatpurple}B})}=\dfrac{\color{flatred}c}{\sin({\color{flatred}C})}\] And thus we have the whole Law of Sines. \(\blacksquare\)

Here's an interactive triangle you can play with. Try out some of the proportions on your calculator, and verify that the proportions match the one shown!

Example

Suppose we have triangle \({\color{flatblue}P}{\color{flatpurple}Q}{\color{flatred}R}\) in the figure below:

Notice that we know both side \(\color{flatred}r\) and angle \(\color{flatred}R\), while we only know angle \(\color{flatpurple}Q\). So, we can set up the Law of Sines and solve for the unknown side \(\color{flatpurple}q\): \[ \begin{align*} \dfrac{\color{flatpurple}q}{\sin({\color{flatpurple}Q})} &= \dfrac{\color{flatred}r}{\sin({\color{flatred}R})}\\ \dfrac{\color{flatpurple}q}{\sin({\color{flatpurple}85^\circ})} &= \dfrac{\color{flatred}7}{\sin({\color{flatred}50^\circ})}\\ {\color{flatpurple}q} &= \dfrac{{\color{flatred}7}\sin({\color{flatpurple}85^\circ})}{\sin({\color{flatred}50^\circ})}\\ {\color{flatpurple}q} &\approx {\color{flatpurple}9.10308}\\ \end{align*} \] We could then find out that \(\color{flatblue}P=45^\circ\) (why?), and set a similar proportion up to find \(\color{flatblue}p\): \[ \begin{align*} \dfrac{\color{flatblue}p}{\sin({\color{flatblue}P})} &= \dfrac{\color{flatred}r}{\sin({\color{flatred}R})}\\ \dfrac{\color{flatblue}p}{\sin({\color{flatblue}45^\circ})} &= \dfrac{\color{flatred}7}{\sin({\color{flatred}50^\circ})}\\ {\color{flatblue}p} &= \dfrac{{\color{flatred}7}\sin({\color{flatblue}45^\circ})}{\sin({\color{flatred}50^\circ})}\\ {\color{flatblue}p} &\approx {\color{flatblue}6.46144}\\ \end{align*} \] Thus we've solved the entire triangle!

---

Preview Activity 12

Answer these questions and submit your answers as a document on Moodle. (Please submit as .docx or .pdf if possible.)

1. Solve the following triangle.
   (That is, find all missing side lengths and angle measures.)
   
2. Solve the following triangle.
   
   How is solving this one different from the previous problem?
3. Answer AT LEAST one of the following questions:
   1. What was something you found interesting about this activity?
   2. What was an "a-ha" moment you had while doing this activity?
   3. What was the muddiest point of this activity for you?
   4. What question(s) do you have about anything you've done?

Answers

---

Return to MAT 130 main page

Posted by Bill Shillito