Lesson Proof of the Law of Cosines revisited

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Law of Cosines

Theorem (Law of Cosines)

For any triangle, with the side lengths a, b, c and opposite angles , and
c%5E2+=+a%5E2+%2Bb%5E2+-+2%2Aa%2Ab%2Acos%28gamma%29

b%5E2+=+a%5E2+%2Bc%5E2+-+2%2Aa%2Ac%2Acos%28beta%29

a%5E2+=+b%5E2+%2Bc%5E2+-+2%2Ab%2Ac%2Acos%28alpha%29

.

Figures 1a) and 1b) illustrate the Theorem, showing an acute and an obtuse triangles.

The side a is opposite to the vertex A and the angle alpha

.
The side b is opposite to the vertex B and the angle beta

.
The side c is opposite to the vertex C and the angle gamma

.
As it is stated, the Theorem is valid for any triangle.

Figure 1a. Acute Triangle

Figure 1b. Obtuse Triangle

The Proof of the Theorem

First, we will prove the theorem for the acute triangle.
Let us draw an altitude of the triangle from one vertex (shown as B) perpendicularly to the opposite side AC.
This altitude BD is shown in red in Figure 2. Triangles ADB and CDB are the right triangles.
Let us denote the segment lengths h=BD, d=AD, e=CD (Figure 2).
From the triangle ADB c%5E2+=+d%5E2%2Bh%5E2

.
From the triangle CDB a%5E2+=+e%5E2%2Bh%5E2

, hence

.
Substitute the last expression to the previous formula for c%5E2

to exclude h. We get c%5E2+=+d%5E2%2Ba%5E2-e%5E2

.
Since

, we have

d%5E2+=+%28b-e%29%5E2+=+b%5E2+-+2%2Ab%2Ae+%2B+e%5E2

.
Substitute this to the previous formula for c%5E2

. We get

c%5E2+=+b%5E2-2%2Ab%2Ae%2Be%5E2%2Ba%5E2-e%5E2+=+a%5E2%2Bb%5E2-2%2Ab%2Ae

.
From the right triangle CDB e+=+a%2Acos%28gamma%29

.
Substitute this to the previous formula for c%5E2

. We get

, which is the first
of the three required equalities. The proof for the remaining two equalities is similar.
So, we proved the theorem for the acute triangle.

Figure 2. Acute Triangle

Now, we will prove the theorem for the obtuse triangle case.
Let gamma

be an acute angle of our triangle. Let us draw an altitude of the triangle from vertex B
perpendicularly to the opposite side AC. This altitude BD is shown in red in Figure 3. Triangles ADB and
CDB are the right triangles. Let us denote the segment lengths h=BD, e=CD (Figure 3).
From the triangle ADB c%5E2+=+%28b%2Be%29%5E2%2Bh%5E2+=+b%5E2%2B2%2Ab%2Ae%2Be%5E2%2Bh%5E2

c%5E2+=+%28b%2Be%29%5E2%2Bh%5E2+=+b%5E2%2B2%2Ab%2Ae%2Be%5E2%2Bh%5E2

.
From the triangle CDB a%5E2+=+e%5E2%2Bh%5E2

, hence

.
Substitute the last expression to the previous formula for c%5E2

to exclude h. We get c%5E2+=+b%5E2%2B2%2Ab%2Ae%2Ba%5E2

.
From the right triangle CDB e+=+a%2Acos%28pi-gamma%29=+-a%2Acos%28gamma%29

.
Note that we used here the equality cos%28pi-gamma%29=-cos%28gamma%29

for the complimentary angle.
Substitute the lasr expression for e to the previous formula for c%5E2

. We get

,
which is the first of three required equality. Since the remaining two angles in our triangle are acute,
the proofs for them are similar to that was done above for the acute triangle.
Thus, we proved the theorem for the obtuse triangle.

Figure 3. Obtuse Triangle

The last remaining case to prove is the right triangle.
If the angle ACB of the triangle ABC is the right angle with the opposite side c, then the Law of Cosines is
c%5E2+=+a%5E2+%2B+b%5E2

,
but this is exactly the Pythagorean theorem. This proves the Law of Cosines validity in this case.
Since the remaining two angles in our triangle are acute, the proofs for them are similar to that was done above for the acute triangle.

Thus, the Law of Cosines Theorem is proved for all cases.

Examples of how to use of the Law of Cosines in solving triangles are given in the lesson Solve triangles using Law of Cosines in this module.

For navigation over the lessons on Properties of Triangles use this file/link Properties of Trianles.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link GEOMETRY - YOUR ONLINE TEXTBOOK.

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