For today’s lesson, we are going to leverage the distance formula and what we already know about right triangle trigonometry to derive the Law of Cosines. The derivation of the formula is not difficult – but I plan to scaffold it so that students see the need for the distance formula and the use of sine and cosine to find an ordered pair.
I start by giving students the task worksheet and have them work on the front side with their table groups. Questions 1 and 2 are very straightforward and students usually have no trouble completing them.
It probably won’t be difficult for students to get the answer to Question 3, but they may find the reference angle of 49.2° and use the right triangle. I challenge them to find the answer with the original angle measure of 130.8°. This will be important because I want them to see that the coordinates of B can be written as (32cos130.8°, 32sin130.8°). This is an important step in the derivation of the Law of Cosines and the most difficult mental leap, so I make a big deal about this so they see the structure.
Question 4 is where it will probably get challenging for most students. If they get stuck, I will try to get them to realize that they could use the distance formula to find the length of segment AB, just like they did for question 1. If a student drops a perpendicular from B to the x-axis to create a right triangle, see if they could find the solution without using that method.
The first question I will ask my students is how they got started to find the length of segment AB for question 4. One they establish that we can use the distance formula, I will ask a student to explain how they found the ordered pairs for points A and B to use in the formula. Then, go through the steps of the distance formula. It is important to not simplify anything while you are working on the board so that students can see the structure of the formula. For example, leave it as (AB)2 = 72 + 62 – 2(6)(7)cos130° so that students can see how the side lengths and the angle measure help us find the last side length of the triangle.
Question 5 moves on to a general case with variables instead of numbers. It is important that students realize that any triangle could be oriented like the triangle in the diagram, so the Law of Cosines will work for all triangle, not just right triangles. I give students a few minutes to use the set up for question 4 as a template to derive the formula with variables instead of numbers. In the video below I discuss this process and how it relates to MP7.
It is vital to discuss what information you need to know about the triangle in order to use the Law of Cosines formula. This is a good time to review SSS and SAS from Geometry and how that information could be plugged into the Law of Cosines to find missing parts. Since these triangles are uniquely determined, there will only be one solution that you could get from the formula. If you knew SSA, then there could be up to two different solutions. I challenge students to think how algebraically you would get two solutions if you were given two sides and a non-included angle and you plugged them into the Law of Cosines.
Question 6 is another example, but has a very different feel since we are finding an angle measure instead of a side length. I give students some time to grapple with this problem. I find that students have the most trouble with deciding which side length should be which in the formula, so pay particular attention to how they set up the problem. In the video below I discuss some common misconceptions about this problem.
To finish the lesson, I want to make sure that my students understand when they can and cannot use the Law of Cosines. As an Exit Ticket, I have my students draw two triangles that each have three pieces of information (side lengths or angle measures). One triangle will be one where the Law of Cosines can be used, for the other the Law of Cosines cannot be used. This is a good indication to me if they understand when this formula will be useful.