Why the Law of Sines Holds
Part of the Triangles Without Right Angles section of Coddy's Trigonometry journey. Lesson 5 of 44.
Drop the height h from C onto the side AB. It cuts the triangle into two right triangles. In the left one, h is opposite A with hypotenuse b, so h = b×sin(A). In the right one, h is opposite B with hypotenuse a, so h = a×sin(B).
Both are the same height: b×sin(A) = a×sin(B). Divide by both sines and you get a/sin(A) = b/sin(B). A height from another corner brings in c the same way. If B is obtuse the height lands outside the triangle, and the right triangle there holds 180° − B, which has the same sine.
Two angles and a side across from one of them (AAS) is the direct case. With A = 40°, B = 75° and a = 9: b/sin(75°) = 9/sin(40°), so b is about 13.52.
Challenge
In triangle ABC, A = 28°, B = 117° and the side a across from A is 9 cm. Find the side x across from B, to two decimal places, then press Check.
Try it yourself
Use the figure to find x
Test yourselfThis lesson includes a short quiz. Start the lesson to answer it and track your progress.
All lessons in Triangles Without Right Angles
2Law of Sines: AAS and ASA
Why the Law of Sines HoldsPair Side With AngleThe Third Angle FirstTwo Coast StationsRecap: Law of Sines8Bearings and Navigation
Reading BearingsTwo Ships From PortTurning Off CourseFinding a Course AngleRecap: Ships at Sea