## Sunday, May 8, 2016

### Laws of Sines Derivation and the Ambiguous Case

I really wanted to make these lessons more meaningful, so I started off with a 3 act math task from the website http://www.101qs.com. This one addressing finding how many pieces of sod cover the triangular garden shown below:
This led to some good questions from the students that made them want to learn how to solve oblique, or non-right triangles.

First, students derived the Law of Sines
Then they practiced a bit. Then they derived the area formula for a triangle using the Law of Sines:
After practicing that a bit, they were able to solve the sod problem...though they got 18 pieces and the video got 20. Hmmmm.

Anyway, the next day we tackled the Ambiguous Case (SSA) for the Law of Sines. We talked about why there could be an ambiguous case for sine and not cosine, i.e., there are two angles less than 180, one that is acute and one that is obtuse for every sinx = a number between 0 and 1. For cosine, if cosx = negative number, the angle would be obtuse and if positive, the angle would be acute.

I told them that there were three possibilities with SSA-two triangles, one, or none. They were reminded of the discriminant at that point! I had each case on a separate piece of paper with scissors, pink paper, rulers, and a glue stick. For each case, they had to draw line segment AB about halfway down the paper, knowing that they did not know it's length (this is also side "c".) They may need to draw it longer or erase from it. But they had to measure the given angle A and cut length "b" from the pink paper. Then they had to cut length "a" to try and have it meet side c (AB). Here's what they did for the first one.

They measured the sides and angles that were missing and then solved by Law of Sines and saw that the answers were very close...and that there was only one possibility for this triangle. When they solved for B, we talked about how B' = 180 - B was the other solution that gave the same sine value. However angle B' + angle A > 180, so there would only be one triangle.

We did this again for case 2, and they saw that no triangle exists, as shown below.
We proved this mathematically, which was pretty cool. I should point out that my sides were a bit long, and next year I will make them smaller...however, students knew that the ratio of the sides just had to be in proportion, so they divided by two in some cases to get them to fit on their paper.

We all had enough of the arts and crafts and for the last case, two solutions, we drew it by hand. It was very cool to see the students grasp the concept of side "a" swinging in to make an oblique angle, and that the two angles, I call them B and B', one acute and one obtuse, both added to 180 degrees. They also saw that angle B' + angle A was less than 180 degrees, so they had room for a third triangle, and therefore there were two possibilities algebraically. Or shall we say, trigonometrically.
They solved for the missing sides of both triangles, and practiced one of each type after that. I felt that they understood the ambiguous case much better.

Tomorrow, we will derive the Law of Cosines and after that, Heron's or Hero's formula.