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anonymous

  • one year ago

DEF m<F=43, d=16mm, and f=24mm, find m<D

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  1. anonymous
    • one year ago
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    43 degrees

  2. jdoe0001
    • one year ago
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    \(\bf \textit{Law of sines} \\ \quad \\ \cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\qquad thus \\ \quad \\ \cfrac{sin(F)}{f}=\cfrac{sin(D)}{d}\implies d\cdot \cfrac{sin(F)}{f}=sin(D) \\ \quad \\ sin^{-1}\left( d\cdot \cfrac{sin(F)}{f} \right)=sin^{-1}[sin(D)]\implies sin^{-1}\left( d\cdot \cfrac{sin(F)}{f} \right)=\measuredangle D\)

  3. anonymous
    • one year ago
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    which is angle A

  4. jdoe0001
    • one year ago
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    hmm that's jus the law of sines notation

  5. anonymous
    • one year ago
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    so what do i plug in for A

  6. jdoe0001
    • one year ago
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    \(\bf \textit{Law of sines} \\ \quad \\ \cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\qquad thus \\ \quad \\ ------------------------------------------------\\ \cfrac{sin(F)}{f}=\cfrac{sin(D)}{d}\implies d\cdot \cfrac{sin(F)}{f}=sin(D) \\ \quad \\ sin^{-1}\left( d\cdot \cfrac{sin(F)}{f} \right)=sin^{-1}[sin(D)]\implies sin^{-1}\left( d\cdot \cfrac{sin(F)}{f} \right)=\measuredangle D \\ \quad \\ \begin{cases} F=43^o\\ f=24\\ d=16 \end{cases}\qquad sin^{-1}\left( 16\cdot \cfrac{sin(43^o)}{24} \right)=\measuredangle D\)

  7. anonymous
    • one year ago
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    um i got 130.

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