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BloomLocke367

  • one year ago

Law of cosines

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  1. BloomLocke367
    • one year ago
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    |dw:1432929186134:dw| I'm looking for angle A.

  2. BloomLocke367
    • one year ago
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    @IrishBoy123

  3. IrishBoy123
    • one year ago
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    |dw:1432929324519:dw| OK :p

  4. BloomLocke367
    • one year ago
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    that's confusing

  5. BloomLocke367
    • one year ago
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    i'm confused

  6. IrishBoy123
    • one year ago
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    @BloomLocke367 for law of cosines, think Pythagoreas Theorem with added power if you want to understand it, that's where to start. if you just want the answer, there will be no limit to the amount of online calculators that will do the work for you. otherwise, you have to walk the long mile.

  7. IrishBoy123
    • one year ago
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    |dw:1432930383845:dw|

  8. BloomLocke367
    • one year ago
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    so what exactly do I do with that?

  9. IrishBoy123
    • one year ago
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    to solve \[a^2 = b^2 + c^2 - 2 \ b \ c \ \cos A\] calculate \[a^2 ,\ b^2, \ c^2 \ and \ 2 bc\] you have all the data cos(A) follows from that

  10. BloomLocke367
    • one year ago
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    so I have289=484+900-1320cosA

  11. BloomLocke367
    • one year ago
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    @IrishBoy123

  12. IrishBoy123
    • one year ago
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    awesome

  13. IrishBoy123
    • one year ago
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    i agree

  14. BloomLocke367
    • one year ago
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    now what do I do?

  15. IrishBoy123
    • one year ago
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    \[1320 \ cosA = 484 + 900 - 289 ;\ \ \ cosA = \frac{484 + 900 - 289}{1320} \]

  16. IrishBoy123
    • one year ago
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    cos A = ?? if it's not 33.948, you can sack me.

  17. BloomLocke367
    • one year ago
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    I got 0.829...

  18. IrishBoy123
    • one year ago
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    good because if cos A = 0.829..., then A = 33.948 [degrees]

  19. BloomLocke367
    • one year ago
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    how so?

  20. BloomLocke367
    • one year ago
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    Can you help with some more?

  21. IrishBoy123
    • one year ago
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    @BloomLocke367 gotta go well done, you can do this stuff easily ciao!

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