8.4. Law of Cosines

To use the law of sines we need a known side opposite a known angle. Sometimes we do not have that information, as when, for example, we know three sides and no angle. We can still solve such a triangle using the law of cosines.

8.4.1. Derivation

Consider an oblique triangle ABC as shown in Fig. 8-27. As we did for the law of sines, we start by dividing the triangle into two right triangles by drawing an altitude h to side AC.

In right triangle ABD,

c2 = h2 + (AD)2

But AD = bCD. Substituting, we get

Now, in right triangle BCD, by the definition of the cosine,


Substituting a cos C for CD in Equation (1) yields

c2 = h2 + (ba cos C)2

Squaring, we have

Figure 8.27. Derivation of the law of cosines.

Let us leave this expression for the moment and write the Pythagorean theorem for the same triangle BCD.

h2 = a2 − (CD)2

Again substituting a cos C for CD, we obtain

Substituting this expression for h2 back into (2), we get

c1 = a2a2cos2C + b2 − 2ab cos ...

Get Technical Mathematics, Sixth Edition now with the O’Reilly learning platform.

O’Reilly members experience live online training, plus books, videos, and digital content from nearly 200 publishers.