![Let (sinA)/(sinB)=(sin(A-C))/(sin(C-B)), where A , B, C are angles of a triangle ABC. If the lengths of the sides opposite these angles are a,b,c respectively, then Let (sinA)/(sinB)=(sin(A-C))/(sin(C-B)), where A , B, C are angles of a triangle ABC. If the lengths of the sides opposite these angles are a,b,c respectively, then](https://d10lpgp6xz60nq.cloudfront.net/web-thumb/121711118_web.png)
Let (sinA)/(sinB)=(sin(A-C))/(sin(C-B)), where A , B, C are angles of a triangle ABC. If the lengths of the sides opposite these angles are a,b,c respectively, then
Derive the formula A = 1/2 ab sin(C) for the area of a triangle by drawing an auxiliary line from a vertex perpendicular to the opposite side - Common Core: High School - Geometry
![Schematic of the AC-SINS assay. Gold nanoparticles coated with capture... | Download Scientific Diagram Schematic of the AC-SINS assay. Gold nanoparticles coated with capture... | Download Scientific Diagram](https://www.researchgate.net/publication/350312286/figure/fig1/AS:1004394422218770@1616477699772/Schematic-of-the-AC-SINS-assay-Gold-nanoparticles-coated-with-capture-antibodies-then.png)