It's time for more triangle shenanigans. The Law of Sines works great when we can use it, but it has a pretty glaring flaw. Like Achilles and his heel, the Law of Sines is vulnerable in its weak point: we have to know at least one of the triangle's angles. No angle, no luck.
That's where the Law of Cosines comes in. Let's look at another triangle to see where it comes from.
The law, not the triangle. Guess we should have been clearer there. We don't really know where the triangle comes from. We just pulled it from our mysterious drawer full of triangles.
When deriving the Law of Sines, we came up with: h = c sin A. In the same way, we can say that f = c cos A, since f is the side adjacent to angle A.
Here's where things take a turn for the weird. Let's focus on the smaller triangle on the right (the one with the sides a, h, and b- f). Why? We're not really sure, we just know it all works out in the end. We tried asking a mathematician, but he just threw down a smoke bomb and ran off in the confusion.
The thing is, we want to know about the sides of the triangle, and we already have an equation for that for the smaller triangle: the Pythagorean Theorem.
h2 + (b - f)2 = a2
We know another way of writing h and f that includes sine and cosine.
(c sin A)2 + (b – c cos A)2 = a2
It's time to work some magic with algebra, starting with the squares. Warning: hairy-looking equation ahead. Get out your trimmers and wade into the thick of it:
c2 sin2 A + (b2 – 2bc cos A + c2 cos2 A) = a2
c2 sin2 A + c2 cos2 A + b2 – 2bc cos A = a2
c2 (sin2 A + cos2 A) + b2 – 2bc cos A = a2
Almost done, then we can try and get rid of all this hair. For the next step, we need the Pythagorean Identity: sin2A + cos2A = 1.
c2(1) + b2 – 2bc cos A = a2
Now, after much wading, we have the Law of Cosines:
a2 = b2 + c2 – 2bc cos A
This works for all three angles/sides in a triangle, so we can also say:
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
We've got 4 variables in these equations: the three sides of the triangle and any one of the angles. As long as we know three of these things, we can suss out the fourth. Yes, we'll suss it.
It'd be a good idea to memorize at least one of these equations. They're kind of important. Here's something to help you out: look at whichever variable is alone. For instance, when it's a2, then the last term, -2bc cos A, doesn't have an a in the "-2bc" part, but it does have an angle A inside the cosine.
Sample Problem
A triangle has sides measuring 4, 7, and 10 units. What is the measurement of the angle opposite the side measuring 7 units?
To start off, we need names for these sides and angles. We like Barkley, Shannon, Lee, and Archibald Williamson the Third, but we should probably pick some more math-y names. How about a = 4, b = 7, c = 10, and B for the angle opposite b? They're rather traditional, but they work.
The Law of Cosines gives the equation:
b2 = a2 + c2 – 2ac cos B
We can start plugging stuff into here and simplifying, so that we can find Archibald Williamson the Third. Er, B.
72 = 42 + 102 – 2(4)(10)cos B
49 = 16 + 100 – 80cos B
-67 = -80cos B
0.8375 = cos B
Now that cos (Archibald Williamson the Third)—sorry, cos B—is isolated, we can find the measurement of B by taking the inverse. Using our calculator in this case.
B = cos-1 (0.8375)
B ≈ 33.1°
Thank you, Law of Cosines, for helping us track down Archibald Williamson the Third. This really is a better name for the angle than B, don't you think? Now that we've found it, it just looks like an "Archibald Williamson the Third."
Sample Problem
A triangle has sides measuring 13 units and 16 units, with an included angle of 42°. What is the length of the side opposite the 42° angle?
We can also use Law of Cosines to find a missing side opposite a known angle, as long as we know the other two sides. Check it:
a2 = b2 + c2 – 2bc cos A
Our mystery side is a, and we have A = 42, b = 13, and c = 16. Let's put those numbers in where they belong:
a2 = 132 + 162 – 2(13)(16)cos 42
All that is left to do here is chug on through, then take a square root to get the value of a:
a2 = 169 + 256 – 416(0.743)
a2 = 115.85
a = 10.8 units
And with that, the mystery is solved. Take that, Sherlock, we don't need you. How about you go back in the van with Shaggy and Scooby.