Both Sine rule and Cosine rule relate the sides of a triangle with its angles. Cosine rule (or Law of Cosines) is
which can also be rewritten as
In triangle , we have , and . Determine .
Applying the cosine rule on , we get that , which gives . Hence, .
*Note: *This is similar to the 'ambiguous case' of Sine rule, since we have , which is the condition .
Suppose and are positive reals such that , and . Show that .
Solution: Since , it follows that and we have similar inequalities for other variables. Hence, the numbers and satisfy the triangle inequality, and there exists a triangle such that .
As such , which gives us that . Similarily, we have , which show that triangle is equilaterial, thus .
Note: This may also be done directly by summing up the 3 equations to get that .
Application and Extensions
Assuming the Cosine Rule, prove the Pythagorean theorem.
In right triangle with , we have , hence , which gives that .
[Apollonius' Theorem] In triangle , is the midpoint of . Show that .
Solution: Apply the cosine rule to triangle , we get . Applying the cosine rule to triangle , we get . Since , so . Adding the first 2 equations, and using the fact that , we get that .
(IMO'68) Determine all triangles whose side lengths are consecutive positive integers, and one of the angles is twice of the other.
Solution: In triangle , let . Let and be the angles at vertices and , respectively. Then .
Applying the cosine rule, we obtain . In a similar manner, we can obtain and . The double angle formula states that . Consider the following cases:
Case 1. . Then, . Clearing denominators and expanding, we get . The only positive integer solution is .
Case 2. . Then, . Clearing denominators and expanding, we get . This has no positive integer solutions.
Case 3. . Then, . Clearing denominators and expanding, we get . The only positive integer solution is . However, checking this triangle, we get the degenerate triangle, hence we ignore this solution.
Thus, the only triangle is the triangle.