Some very cool things about the Law of Sines

So yeah, I’m teaching the law of sines and cosines, and I’m finding some awesome things. What’s totally ridiculous is that after I introduced some of it to my class, I looked back at my stuff from last year and apparently I had done the same thing. Like… I don’t remember it at all.

In any case, when I teach the Law of Sines, I tend to have kids derive it by finding the area of a triangle in three different ways.

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We set these three different ways to get the area of a triangle equal to each other, and divide by \frac{1}{2}abc to get the Law of Sines:

\frac{\frac{1}{2}ba\sin(C)}{\frac{1}{2}abc}=\frac{\frac{1}{2}cb\sin(A)}{\frac{1}{2}abc}=\frac{\frac{1}{2}ac\sin(B)}{\frac{1}{2}abc}

\frac{\sin(C)}{c}=\frac{\sin(A)}{a}=\frac{\sin(B)}{b}

Now, to be clear, there are some subtleties that have to be addressed here. Like for example, this argument clearly works for acute triangles, but what about an obtuse triangle or right triangle, like:

 

It turns out with just a tiiiiny little bit of extra work, we can show that the Law of Sines holds. (Here’s a fun little applet you can play with for this… one important thing that can help you for the obtuse triangle proof is that \sin(\theta)=\sin(180-\theta).)

So… yeah. That’s a pretty traditional way to teach the Law of Sines. But did you know that the ratio that pops up with the Law of Sines has a geometric interpretation?

Like, look at this triangle. And look at the ratio of (side length)/(the sine of the angle opposite the side).

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That 10.36 has a geometric meaning. Ready for it? READY? I don’t think you are, but I’m going to show it to you anyway…

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Dang! HOLY MOTHERFATHER! Yuppers. That triangle has one circle that can perfectly circumscribe it. And twice the radius of that circumscribed circle is that ratio!!! Don’t believe me? Ok, I know you do, but play with this applet I made to see it happen! Maybe try to create a right triangle and see if that reminds you of something you learned in geometry?

Now this year I told my kids that \frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}=2R. And I sent ’em up to the whiteboards and asked them to prove it. I gave some hints. Like, for example, the inscribed angle theorem. But eventually kids got it!

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This is actually another proof of the Law of Sines! (To be clear, you will also have to make an argument for an obtuse triangle, which requires a tiny bit of modification. You have to see a central angle and one of the angles of the triangle are the same because they both subtend the same arc. And a right triangle.)

So I had a follow up after this… I asked kids to prove that the area of any triangle is: \frac{abc}{4R}, where R is the radius of the circumscribed circle. I asked them to prove it algebraically, and they did:

\text{Area}=\frac{1}{2}ab\sin(C). But we know that \frac{c}{\sin(C}=2R. So let’s manipulate the rea equation to get an R in it.

\text{Area}=\frac{1}{2}abc\frac{\sin(C)}{c}.

Now we have \text{Area}=\frac{1}{2}abc\frac{1}{2R}=\frac{abc}{4R}.

I asked kids to show this algebraically. They did it in various ways (all correct), similar to the argument above. However I had a student present me with a stunning geometric argument that proved this area formula. I honestly don’t know if I would have been able to come up with it. It was so stunning I had to take a photo of it. I leave this as an exercise for the reader. MWAHAHAHA.

(All of this Law of Sine stuff was inspired by this webpage.)

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6 comments

  1. One question about the first part: did you justify before the very notion of area of a triangle? I mean: why does the three different expressions for the triangle area produce the same value? Because that’s the basis for your proof of the Law. Of course, one possible answer is triangle similarity, which is also the reason why the Law is true. I’m truly curious about your approach.

    1. I’m not sure I totally understand your question… The three area formulas for the triangle come from doing (1/2)(base)(height), but just by using the three different sides as bases… So since they are the area for the same triangle, calculated three different ways, they all must be equal.

      1. Maybe I’m just not getting what you’re saying. They all are the same because they represent the area of the same object (the triangle).

  2. Pretend you don’t know that for a moment. When you learned the formula for the area of a triangle, (1/2)(base)(height), didn’t you ask why that formula worked whatever the base(and the height)?

    1. Hihi! Okay! I see (I think?)! I taught these kids in geometry, where we’ve proved that formula before (in a few different ways). So that’s why I didn’t ask. Here is one of them that I especially enjoyed when kids thought about it!

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