The Law of Sines can be very useful, but there are cases where it won't do. Don't get us wrong. It's taken us this far and we owe it some gratitude. But no man is an island and the Law of Sines can't do it all alone. Take this triangle, for example.
Any way we try, the Law of Sines just won't cut it. Luckily, we have another law that we can use. Since this second law cooperates well with the Law of Sines, we'll call it the Law of Cosines.
For any triangle, represented by the triangle above, the Law of Cosines says that the following are true:
a^{2} = b^{2} + c^{2} – 2bc cos A
b^{2} = a^{2} + c^{2} – 2ac cos B
c^{2} = a^{2} + b^{2} – 2ab cos C
These formulas will help us when the Law of Sines can't, but before we use them, we have to prove them.
We'll start off with a slightly altered Pythagorean theorem. Let AD = x, which means that DB = c – x. In this case, a and b are the hypotenuses.
(c – x)^{2} + h^{2} = a^{2}  Apply Pythagorean theorem to ΔCDB 
c^{2} – 2cx + x^{2} + h^{2} = a^{2}  Multiplying out (c – x)^{2} 
x^{2} + h^{2} = b^{2}  Apply Pythagorean theorem to ΔADC 
c^{2} – 2cx + b^{2} = a^{2}  Substitute in b^{2} for x^{2} + h^{2} 
cos A = ^{x}⁄_{b}  Definition of cos A 
b cos A = x  Rearrange (multiply both sides by b) 
c^{2} – 2bc cos A + b^{2} = a^{2}  Substitute in b cos A for x 
a^{2} = b^{2} + c^{2} – 2bc cos A  Rearrange, thanks to the commutative property 
We're good to go. Bring it on, triangles.
Now we can solve this triangle. Let the only unknown side be a, ∠A = 84°, and b = 10 and c = 12. We'll start with our fresh new law.
a^{2} = b^{2} + c^{2} – 2bc cos A
Fill it up.
a^{2} = (10)^{2} + (12)^{2} – 2(10)(12) cos(84°)
a^{2} = 100 + 144 – 240 cos(84°)
a^{2} = 244 – 25.1
a^{2} = 218.9
a ≈ 14.8
Great. If we wanted to solve the entire triangle, we could find the remaining angles using either the Law of Sines or the Law of Cosines. Hooray for freedom.
Law of Sines is also useless if we have a triangle without any known angles.
With this triangle, the Law of Sines wouldn't amount to squat. Zip. Zero. Nada. Thank goodness we have the Law of Cosines to help us out.
a^{2} = b^{2} + c^{2} – 2bc cos A
Hold up. If we plug in all the sides now, moving the terms around will be more uncomfortable than an elephant in a broom closet. Isolating our variable will make things easier for us in the long run. When we do, we end up with this piece of work:
Now we can substitute our values without breaking a sweat.
All right. Time to pull out the abacus. On second thought, a calculator is probably a better idea.
A ≈ 87.2°
Nice job. Since we have an angle, we can choose whether to use the Law of Cosines or the Law of Sines to solve for the other two angles. The Law of Sines is probably easier, but that's not what this chapter's all about.
b^{2} = a^{2} + c^{2} – 2ac cos B
Much like the Power Rangers, it's morphin' time!
Now we can plug in our side lengths.
B ≈ 55.4°
Wonderfully done. For the last angle, we have not one, not two, but three different options.
We've done them all, and they'll all give us the same answer. Just to practice, we'll give the Law of Cosines one more go.
c^{2} = a^{2} + b^{2} – 2ab cos C
We're still looking for an angle. Let's act like it.
Replace the letters with numbers. This is math, after all.
C ≈ 37.4°
The Law of Cosines, while very useful, is also very jumbled. All the letters and numbers and exponents and pluses and minuses… it's easy to get confused. The Law of Sines on the other hand, is quite simple. That's why with the Law of Cosines, it's best to cosign with some other formulas so that things don't get too muddled. Cosine…cosign…get it?
Find the value of x.

A triangle has side lengths of 10, 11, and 12. Find all the angles. 
Find the value of x, y, and z.

Find the measurements of all angles in the triangle.
Solve the triangle.
Find ∠R if ∆QRS has the following points: Q (3, 1), R (1, 4), and S (1, 2).
Find the value of x.
Find the length of AD if ∠BCE = 180° exactly.
In hopes to lure Jerry out and catch him, Tom has planted a piece of cheese at a distance away from his mouse hole. Tom is ready to pounce from his hiding spot, 7 feet away from the mouse hole and 10 feet away from the cheese. If the angle Tom makes is 73.1°, approximately how far away did he place the cheese?
A tilted tree stands at a height of 30.6 feet. The shadow it casts is 37.8 feet long, 12.1 feet past a perpendicular line from top of the tree to the ground. If the tree is actually 32.9 feet tall, what angle does the tree make with the ground?