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anonymous
 one year ago
DEF m<F=43, d=16mm, and f=24mm, find m<D
anonymous
 one year ago
DEF m<F=43, d=16mm, and f=24mm, find m<D

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf \textit{Law of sines} \\ \quad \\ \cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\qquad thus \\ \quad \\ \cfrac{sin(F)}{f}=\cfrac{sin(D)}{d}\implies d\cdot \cfrac{sin(F)}{f}=sin(D) \\ \quad \\ sin^{1}\left( d\cdot \cfrac{sin(F)}{f} \right)=sin^{1}[sin(D)]\implies sin^{1}\left( d\cdot \cfrac{sin(F)}{f} \right)=\measuredangle D\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm that's jus the law of sines notation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so what do i plug in for A

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf \textit{Law of sines} \\ \quad \\ \cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\qquad thus \\ \quad \\ \\ \cfrac{sin(F)}{f}=\cfrac{sin(D)}{d}\implies d\cdot \cfrac{sin(F)}{f}=sin(D) \\ \quad \\ sin^{1}\left( d\cdot \cfrac{sin(F)}{f} \right)=sin^{1}[sin(D)]\implies sin^{1}\left( d\cdot \cfrac{sin(F)}{f} \right)=\measuredangle D \\ \quad \\ \begin{cases} F=43^o\\ f=24\\ d=16 \end{cases}\qquad sin^{1}\left( 16\cdot \cfrac{sin(43^o)}{24} \right)=\measuredangle D\)
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