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schmidtdancer

  • 3 years ago

Please help fast, I will give medal!! Use the given information to solve the triangle. A=60 degrees, a=9, c=10. This is an ambigous case, so there needs to be 2 triangles!! thanks!!

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  1. schmidtdancer
    • 3 years ago
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    the first triangle is A=60 deg, B=46 deg, C=74 deg, sides a=9, b=7.4, c=10. I need help with the other onee!! please

  2. colorful
    • 3 years ago
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    A is the angle between a and c or what?

  3. schmidtdancer
    • 3 years ago
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    it doesn't say..

  4. colorful
    • 3 years ago
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    well there must be some picture or something that indicates whether c is touching angle A or not, otherwise all we have is...

  5. schmidtdancer
    • 3 years ago
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    There isnt a picture, i promise

  6. colorful
    • 3 years ago
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    oh I bet you are supposed to assume angle A is opposite side a, which makes this doable with the law of sines I think

  7. schmidtdancer
    • 3 years ago
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    yeah

  8. colorful
    • 3 years ago
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    |dw:1340038642500:dw|

  9. schmidtdancer
    • 3 years ago
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    . Find the bearing of the flight from C to A. a plane flies 500 km with a bearing of N 44 degrees west from B to C. the plane then flies 720 km from c to a. find thhe bearing of the flight from c to a? can you do this one too

  10. colorful
    • 3 years ago
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    how 'bout we do one at a time ;)

  11. schmidtdancer
    • 3 years ago
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    Ok

  12. colorful
    • 3 years ago
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    first get all three angles of the triangle\[\frac a{\sin A}=\frac c{\sin C}\]solve for C then you know that B=180-A-C, so you will have all 3 angles, which will allow you to find the third side b\[\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}\]

  13. schmidtdancer
    • 3 years ago
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    Okay I got that

  14. schmidtdancer
    • 3 years ago
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    so....

  15. momo0000
    • 3 years ago
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    SINA/a=SINB/b=SINC/c SIN60/9=SINC/10. SOLVING= 10SIN60/9=C 180-A+C=B then sinBsinA/a=b

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