The Law of Cosines Study Guide
The Law of Cosines
When we know the angles of a triangle and one of the sides, we can figure out the lengths of the other sides with the Law of Sines. In this lesson, we see how a new theorem, the Law of Cosines, enables us to find the third length of a triangle when we know the lengths of the other two sides and the angle between them. For example, suppose we know the lengths A and B of the triangle in Figure 18.1 and the measure θ of the angle between them. We would like to find the length C of the third side.
Law of Cosines
C2 = A2 + B2 – 2AB·cosθ
We can prove this with vectors (if the proof is confusing, feel free to skip to the examples on page 175). To make things easier, rotate the triangle so that the angle θ is in standard position, as shown in Figure 18.2.
Let the origin be the right endpoint of the side with length B. If we look at the side of length B as a vector, then its components are (–B,0), because it goes a distance of B to the left and does not go up or down at all. The side with length A goes up at an angle of θ , so as a vector, its components are (A· cos(θ),A · sin(θ)). The sum of these two vectors is thus (A· cos(θ) – B,A· sin(θ)), as illustrated in Figure 18.3.
The side with length C goes from the origin (0,0) to the point (A· cos(θ) – B,A· sin(θ)). This means that it is the hypotenuse of a triangle with legs of length | A · cos(θ) – B | and A · sin(θ), as shown in Figure 18.4. Because we don't know if A · cos(θ) or B is bigger, we use absolute values to describe the length of the bottom of this triangle. When we square this length, it won't matter whether A · cos(θ) – B was positive or negative.
We can now find the length C with the Pythagorean theorem:
- C2 = (A· sin(θ))2 + (A· cos(θ) – B)2
- C2 = A2· sin2(θ) + A2 · cos2(θ) – 2AB · cos(θ) + B2
We can factor the A2out of the first two terms on the right.
- C2 = A2(sin2(θ) + cos2(θ)) – 2AB · cos(θ) + B2
If we use sin2(θ) + cos2(θ) = 1, proved in Lesson 6, we end up with the Law of Cosines:
- C2 = A2 – 2AB · cos(θ) + B2
Usually, this is written as C2 = A2 + B2 – 2AB · cos(θ). We can use this to quickly find the lengths of triangles.
Find the length x in Figure 18.5.
We can use the Law of Cosines with A = 8, B = 11, C = x, and θ = 39°:
- C2 = A2 + B2 – 2AB · cos(θ)
- x2 = 82 + 112 – 2(8)(11) · cos(39°) ≈ 48.22
- x ≈ √48.22 ≈ 6.9
The length is x ≈ 6.9 inches.
Find the length x in Figure 18.6.
We use the Law of Cosines with A = 40, B = 72, C = x, and θ = 108°.
- C2 = A2 + B2 – 2AB · cos(θ)
- x2 = 402 + 722 – 2(40)(72) · cos(108°) ≈ 8,563.94
- x ≈ √8,563.94 ≈ 92.5 cm
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