how to derive cosine law from sine law? can anybody help me.. please..

Question

anonymous · Answer

the law of sines tells us that for a triangle with sides $a,b,c$ and opposite angles $A,B,C$ we have $$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}c=K$$ for some constant $K$

anonymous · Answer

now consider $A+B+C=\pi$ so expand: $$\sin(\pi-B-C)=\dots\\sin(\pi-A-C)=\dots\\sin(\pi-A-B)$$

no.identity05 · Answer

i really don't get your point, can you show me how?
this formulas must be use (a/sinA)=(b/sinB)=(c/sinC)=2R
where R is raduis of circumcircle
and area of triangle=(1/2)(ab)sinC

loser66 · Answer

https://en.wikipedia.org/wiki/Law_of_cosines Read the part "using law of sin", it is exactly what @oldrin.bataku wrote

no.identity05 · Answer

thanks mate..