Question

DEF m 10 years ago

Answer 1

43 degrees

Answer 2

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

Answer 3

which is angle A

Answer 4

hmm that's jus the law of sines notation

Answer 5

so what do i plug in for A

Answer 6

\(\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\)

Answer 7

um i got 130.