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adnanchowdhury

  • 2 years ago

Trigonometry question on R cos (theta - alpha). http://d.pr/L96v

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  1. adnanchowdhury
    • 2 years ago
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    Second part of the question is what i need help with. My answer to the first part: \[\sqrt{17} \cos (\Theta + \alpha)\] Second part: My answers are: 46.9 and -46.9. But answers says: 46.9 and -75?

  2. adnanchowdhury
    • 2 years ago
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    Edit: I mean to write, the answer to my first answer was √cos(Θ+14)

  3. satellite73
    • 2 years ago
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    [r\sin(x+\theta)\] where \[r=\sqrt{a^2+b^2}, \tan(\theta)=\frac{a}{b}\]if i recall correctly

  4. satellite73
    • 2 years ago
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    oops you want cosine, sorry

  5. satellite73
    • 2 years ago
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    same idea but this time \[r\cos(x - \theta)\] same r but \[\tan(\theta)=\frac{b}{a}\]

  6. satellite73
    • 2 years ago
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    so we need \[\theta\] in degrees i take it, yes?

  7. adnanchowdhury
    • 2 years ago
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    yes

  8. satellite73
    • 2 years ago
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    i get \[\theta = -14.4\] giving \[\sqrt{17}\cos(x+14.4)\]

  9. satellite73
    • 2 years ago
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    \[\sqrt{17}\cos(x+14.4)=2\] \[\cos(x+14.4)=\frac{2}{\sqrt{17}}\] \[x=\cos^{-1}(\frac{2}{\sqrt{17}})-14.4\]

  10. adnanchowdhury
    • 2 years ago
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    between -180 and 180 degrees.

  11. adnanchowdhury
    • 2 years ago
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    Mark scheme: http://d.pr/ubFq Where does the value -75 come from?

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