![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/647742846_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
![geometry - Understanding this proof for $\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$ from Gelfand - Mathematics Stack Exchange geometry - Understanding this proof for $\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$ from Gelfand - Mathematics Stack Exchange](https://i.stack.imgur.com/H1tNx.png)
geometry - Understanding this proof for $\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$ from Gelfand - Mathematics Stack Exchange
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Schematic of the AC-SINS assay. Gold nanoparticles coated with capture... | Download Scientific Diagram
![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)