• Solve oblique triangles using the law of cosines (SAS,SSS)
• Decide which law (Sines or Cosines) to use in solving an oblique triangle

The law of cosines can be used when two sides and the included angle are known or when all three sides are known. This is the situation with SAS and SSS.

Note that a² = b² + c² - 2bc cos A simplifies somewhat when the measure of angle A is 90°. Then cos A = cos 90° = 0 so the Law of Cosines becomes a² = b² + c² (the pythagorean theorem). In an acute triangle, a² < b² + c² and in an obtuse triangle, a² > b² + c²

Example 1 (SSS):

In triangle XYZ, x = 3, y = 7 and z = 9. Find the measure of angle Z.

z² = x² + y² - 2xy cos Z
Substitute and solve.

Example 2 (SSS):

In triangle XYZ, x = 3, y = 7 and z = 11. Find the measure of angle Z.

z² = x² + y² - 2xy cos Z
Substitute and we have trouble--what is it?

Example 3 (SAS):

In triangle ABC, b = 4, c = 5 and the measure of angle A is 51°. Solve the triangle. Can you use the Law of Sines first?

Example 4 (Extra for experts):

You can use the law of cosines to solve the ambiguous case that we looked at with the law of sines. The trick is to write the law of cosines as a quadratic in terms of the unknown side. In our last example with the law of sines, triangle ABC had angle B and sides a and b given. So we use:

b² = a² + c² - 2ac cos B

or, rearranged:

c² + (-2acosB)c + (a² - b²) = 0

Plug that into the quadratic formula and:

• If the discriminant b² - a²sin²B < 0 , then there is no solution (roots are complex). Simplified, we get b < a sinB.
• If the discriminant b² - a²sin²B = 0, then there is one solution. Simplified, we get b = a sinB.
• If the discriminant b² - a²sin²B > 0, then there are two solutions (b > a sinB) until b > a, at which point there is a single solution.
• For example, with a = 5, b = 2, and angle B = 27°, we get:
  2 = b < a sinB = (5)(sin(27°) = 2.2699524986977339578020418317894 so there is no solution since the roots for c are complex.