Law of Sines (Lesson 5-5) The Law of Sines is an extended proportion. Each ratio in the proportion is the ratio of an angle of a triangle to the length of its opposite side. Law of Sines: sin A/a = sin B/b = sin C/c where a is the opposite side of angle A, b is the opposite side of angle B, and c is the opposite side of angle C

Solving the Proportion If the known parts of the triangle are AAS or ASA, then you can solve the Law of Sines proportion for the other three values not known. Example 1, pg. 479

Solving when SSA is known. If you are given m∠A, a, and b, then there is the possibility that no triangle exists, one triangle exists, or two triangles exist.

Solving when SSA is known. Use these steps, when SSA is known: 1) If A is obtuse and a > b, then there is one triangle (use Law of Sines to solve). 2) If A is obtuse and b ≥ a, then there is no triangle that exists for these measures. 3) If A is acute and a ≥ b, then there is one triangle (use Law of Sines to solve). 4) If A is acute and sin A = a/b, then there is one triangle, (use Law of Sines to solve). 5) If A is acute and b · sin A > a, then there is no triangle that exists for these measures. 6) If none of the above are satisfied, then there are two triangles (This is called the ambiguous case). a) Find m∠B1 using the Law of Sines and continue solving for C1 and c1. b) For the second triangle B2 = 180 – B1, then continue solving for C2 and c2.

Example 2, pg. 480

Example 3, pg. 481

Homework Pages 484, #2-20 even.