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meg1334

  • one year ago

will give medal to best answer! The given measurements may or may not determine a triangle. If not, then state that no triangle is formed. If a triangle is formed, then use the Law of Sines to solve the triangle, if it is possible, or state that the Law of Sines cannot be used. C = 38°, a = 19, c = 10

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  1. Hero
    • one year ago
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    Did you try simply using Law of Sines to figure it out?

  2. meg1334
    • one year ago
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    yes but i didnt understand

  3. Hero
    • one year ago
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    What did you come up with?

  4. meg1334
    • one year ago
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    i got sin A = 67 but that isnt in my answer choices No triangle is formed. A = 58.6°, B = 83.4°, b ≈ 15.6 A = 83.4°, B = 58.6°, b ≈ 15.6 The triangle cannot be solved with the Law of Sines.

  5. Hero
    • one year ago
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    I don't understand how you got that. Please show me the steps you took to get A = 67

  6. meg1334
    • one year ago
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    i used c/ sin c = a/ sin a 10/sin 37 = 19/sin a and then i divided 19 sin 38 over sin 10 and i got 67.36

  7. Hero
    • one year ago
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    where did you get the 37 from?

  8. meg1334
    • one year ago
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    I meant 38, i used 38 in my claculations and still got 67

  9. meg1334
    • one year ago
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    calculations

  10. Hero
    • one year ago
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    Are you in degree mode or radian mode?

  11. meg1334
    • one year ago
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    i dont know i just typed it in and got 67, cant you check it?

  12. Hero
    • one year ago
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    You must make sure you're in degree mode. That's very important.

  13. Hero
    • one year ago
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    But either way it goes, you're not doing it correctly.

  14. meg1334
    • one year ago
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    oh really, i hadnt noticed that im doing it wrong. I need help thats why im on here obviously

  15. Hero
    • one year ago
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    \[\sin A = \frac{19\sin(38^{\circ})}{10} \\\sin A = \]

  16. Hero
    • one year ago
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    Compute the right hand side again and let me know what you get. Make sure you're in degree mode.

  17. meg1334
    • one year ago
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    1.169756803

  18. Hero
    • one year ago
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    Okay now \[\sin A = 1.169756803\] So how do we find A?

  19. meg1334
    • one year ago
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    a is 19, sin A is 1.16, what do you mean find A?

  20. Hero
    • one year ago
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    A is an angle. Find it

  21. Hero
    • one year ago
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    A is an Angle like theta

  22. Hero
    • one year ago
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    I'm not talking about a = 19. That's a side length

  23. Hero
    • one year ago
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    \[\sin (A^{\circ}) = 1.169756803\] I'm talking about Angle A. Find that.

  24. meg1334
    • one year ago
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    how?

  25. Hero
    • one year ago
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    You take the inverse sine of both sides to get: \[\sin^{-1}(A^{\circ}) = \sin^{-1}({1.169756803}) \\A^{\circ} = \sin^{-1}({1.169756803}) \]

  26. Hero
    • one year ago
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    So compute \[\sin^{-1}({1.169756803})\]

  27. Hero
    • one year ago
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    Let me know what you get.

  28. meg1334
    • one year ago
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    it says error

  29. Hero
    • one year ago
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    Exactly.

  30. Hero
    • one year ago
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    That's what it should say because no such angle exists.

  31. Hero
    • one year ago
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    And if the angle doesn't exist, then the triangle can't possibly exist.

  32. Hero
    • one year ago
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    \[-1< \sin(\theta) < 1\] means that the value of sine can only be between - or + one. So there's no such thing as \(\sin(\theta) = 1.17\)

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