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no.identity05

  • one year ago

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

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  1. anonymous
    • one year ago
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    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 \(K\)

  2. anonymous
    • one year ago
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    now consider \(A+B+C=\pi\) so expand: $$\sin(\pi-B-C)=\dots\\\sin(\pi-A-C)=\dots\\\sin(\pi-A-B)$$

  3. no.identity05
    • one year ago
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    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

  4. Loser66
    • one year ago
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    https://en.wikipedia.org/wiki/Law_of_cosines Read the part "using law of sin", it is exactly what @oldrin.bataku wrote

  5. no.identity05
    • one year ago
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    thanks mate..

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