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BloomLocke367

  • one year ago

In FGH, FH=9, FG=11 and angle F=65. What is the measure of angle G?

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  1. BloomLocke367
    • one year ago
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    it's the same process of last time, right?

  2. BloomLocke367
    • one year ago
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    law of sines?

  3. geerky42
    • one year ago
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    |dw:1433257580394:dw| I don't think so. No pair of angle and its opposite side to use.

  4. geerky42
    • one year ago
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    let me think for a miin

  5. BloomLocke367
    • one year ago
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    okay

  6. geerky42
    • one year ago
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    I believe way to do it is to use law of cosines first, to solve for GH, then from there, we can use law of sines to solve for G. Good plan?

  7. geerky42
    • one year ago
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    \(g^2+h^2-2gh\cos F = f^2\)

  8. BloomLocke367
    • one year ago
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    I guess so.

  9. BloomLocke367
    • one year ago
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    give me some time to work that out

  10. BloomLocke367
    • one year ago
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    so far I have \(10\cos(65)=f^2\), is that right so far?

  11. geerky42
    • one year ago
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    I have \(9^2+11^2-2(9)(11)\cos(65^\text o) = f^2\)

  12. BloomLocke367
    • one year ago
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    yes, then it's 81+121-192cos(65), right?

  13. geerky42
    • one year ago
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    81+121-198cos(65)

  14. geerky42
    • one year ago
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    Yes

  15. BloomLocke367
    • one year ago
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    okay, so now I have 10cos(65)=f^2. now what?

  16. geerky42
    • one year ago
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    that number next to cos, it's 198, not 192. And you don't add number together; \(a^2+b^2-2ab\cos\theta~~~\neq~~~(a^2+b^2-2ab)\cos\theta\)

  17. BloomLocke367
    • one year ago
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    I wrote it down wrong from my calculator XD thanks

  18. BloomLocke367
    • one year ago
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    so it's 4, not 10

  19. geerky42
    • one year ago
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    \[\large a^2+b^2-2ab\cos\theta~~~\neq~~~(a^2+b^2-2ab)\cos\theta\] When calculating -2ab part, you need to multiply it by cos first before add them to other numbers

  20. BloomLocke367
    • one year ago
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    makes sense, order of operations. sorry.

  21. geerky42
    • one year ago
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    just completely calculate left side, what do you get?

  22. BloomLocke367
    • one year ago
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    118.32

  23. BloomLocke367
    • one year ago
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    @geerky42

  24. BloomLocke367
    • one year ago
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    so the angle is 10.9

  25. geerky42
    • one year ago
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    actually it's side opposite of angle F. it's 10.9 .

  26. geerky42
    • one year ago
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    WE calculated for side f.

  27. BloomLocke367
    • one year ago
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    ohhhhh, duh. XD

  28. BloomLocke367
    • one year ago
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    sorry XD

  29. geerky42
    • one year ago
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    NOW we can use law of sine here. \(\dfrac{\sin F}{f} = \dfrac{\sin G}{g}\)

  30. BloomLocke367
    • one year ago
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    sin65/10.9=sin(G)/9

  31. BloomLocke367
    • one year ago
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    9sin(65)=10.9sin(G)

  32. geerky42
    • one year ago
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    Or you can just multiply both sides y 9

  33. BloomLocke367
    • one year ago
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    \(\Large\frac{9\sin(65)}{10.9}=\sin(G)\)

  34. geerky42
    • one year ago
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    Yeah

  35. BloomLocke367
    • one year ago
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    then multiply it by arcsin, right?

  36. BloomLocke367
    • one year ago
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    I got 48.45

  37. geerky42
    • one year ago
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    right \[\arcsin\left(\frac{9\sin(65)}{10.9}\right)=G\]

  38. geerky42
    • one year ago
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    Same here

  39. BloomLocke367
    • one year ago
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    awesome, thanks for your help? Help with more? (sorry, I know I must be bothering you)

  40. geerky42
    • one year ago
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    I need to go soon. One more.

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