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milliex51

How do I find the measure of the indicated angle, to the nearest tenth of a degree using sine law and cosine law?

  • one year ago
  • one year ago

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  1. milliex51
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    |dw:1337042419972:dw|

    • one year ago
  2. jiberjiber
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    Uh, what?

    • one year ago
  3. jiberjiber
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    |dw:1337043771132:dw|

    • one year ago
  4. jiberjiber
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    four degrees?

    • one year ago
  5. milliex51
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    no, it's 104 degrees and the side length of BC is 45 m

    • one year ago
  6. Mertsj
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    |dw:1337046316750:dw|

    • one year ago
  7. Mertsj
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    Let's say that side AB is c and use the law of cosines.

    • one year ago
  8. milliex51
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    |dw:1337046422125:dw| I found the side length of AC using cosine law: AC^2 = BC^2 + AB^2 - 2(BC)(AB)cosB = 45^2 + 20^2 - 2(45)(20)cos104 = 2860.459412 AC = 53.48 m

    • one year ago
  9. Mertsj
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    \[c^2=20^2+45^2-2(20)(45)\cos 104\]

    • one year ago
  10. milliex51
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    and then I used cosine law again to find the side length of DC: DC^2 = BC^2 _ BD^2 -2(BC)(BD)cosB = 45^2 + 30^2 -2(45)(30)cos(104 degrees) = 3578.19 I used the triangle BCD

    • one year ago
  11. milliex51
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    and now I am stuck..

    • one year ago
  12. Mertsj
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    |dw:1337046557115:dw|

    • one year ago
  13. Mertsj
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    Now find the meassure of angle ACB using the law of sines.

    • one year ago
  14. Mertsj
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    |dw:1337046713091:dw|

    • one year ago
  15. milliex51
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    but what angle am I suppose to find for triangle ABC, when angle B=104 degrees? wait I am confused..

    • one year ago
  16. Mertsj
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    \[\frac{\sin 104}{53.48}=\frac{\sin ACB}{20}\]

    • one year ago
  17. Mertsj
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    Angle ACB = 21.3 degrees

    • one year ago
  18. milliex51
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    how did you get 21.3?

    • one year ago
  19. milliex51
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    and the angle of C has to be 7.3 degrees..

    • one year ago
  20. Mertsj
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    |dw:1337046878417:dw|

    • one year ago
  21. Mertsj
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    20(sin 104)/52.48 Then hit the inverse sin button.

    • one year ago
  22. milliex51
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    how did you get 54.7 and 21.3? :O i'm so sorry

    • one year ago
  23. Mertsj
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    Look above. It tells you how to get the 21.3. Then use the fact that the sum of the angles of a triangle is 180 to get 54.7

    • one year ago
  24. milliex51
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    ohh yup. But how am I going to find angle C? D:

    • one year ago
  25. Mertsj
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    But you want the angle theta so try this: \[(CD)^2=30^2+45^2-2(30)(45)\cos 104\]

    • one year ago
  26. Mertsj
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    |dw:1337047600966:dw|

    • one year ago
  27. milliex51
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    i'm sorry but it just says, 'math processing error..'

    • one year ago
  28. Mertsj
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    |dw:1337047693118:dw|

    • one year ago
  29. Mertsj
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    Now use the law of sines to find angle DCB. Subtract 21.3 from DCB and that will be theta.

    • one year ago
  30. milliex51
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    okay. I'm still kind of confused but thank you so much!

    • one year ago
  31. Mertsj
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    \[\frac{\sin 104}{59.8}=\frac{\sin DCB}{30}\]

    • one year ago
  32. Mertsj
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    What are you confused about?

    • one year ago
  33. milliex51
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    Mertsj, how come the answer at the back fo my book is 7.8 degrees? o.o

    • one year ago
  34. milliex51
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    ohh, never mind :$ THANK YOU!!

    • one year ago
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