# Thinking!

I love it when my students think for themselves.

When learning the law of sines and law of cosines — and when it’s appropriate to use one or the other — our textbook gives pretty prescriptive directions. For example, when you are given SSA (a side, a side, and an angle opposite one of those given sides), you are supposed to use the law of sines. And depending on these values, you can actually get two possible solutions!

Let’s work this out.

If in triangle ABC you’re given that side a=6, side b=8, and angle A is 40 degrees, then let’s solve the triangle.

Using the law of sines, $\frac{6}{\sin (40)}=\frac{8}{\sin B}$. Rearranging, we get $\sin B=0.8571$. But we know that sine is positive in quadrants I and II, so we get for $B=58.99$ or $B=121.01$. And hence, we have two possible angle values for B, which leads to two different triangles. [For a more detailed explanation, see here.] Another quick application of the law of sines yields that $c=9.22$ or $c=3.04$

And looking at the picture below (cribbed from the site above), you can see that both triangles are possible!

Here’s where student thinking is awesome. The book, as I said, says that everytime you have SSA you should use the law of sines. And I agree, it is easier. But it is definitely possible to use the law of cosines too, as one of my students pointed out to me.

Let’s do it:

$6^2=8^2+c^2-2(8)(c)\cos(40)$

This simplifies, with some rearranging, to the quadratic: $c^2-12.26c+28=0$. This can be solved to get the two values for $c$, which are $c=9.22$ or $c=3.04$.

I love that it works, and that the student insisted that we could do it. It might be slightly more work, but not that much more, and the exploration aspect is awesome. [1]

[1] An extension for a project for next year might be: how can we use the fact that we can generate a quadratic help us in determining when we have two possible triangles, one possible triangle, or no possible triangles.