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Pepsi [2]
3 years ago
5

How would I find a given triangle where: A=0.1, B=1 and c=9? Thanks in advance.

Mathematics
1 answer:
Juliette [100K]3 years ago
3 0
\bf \textit{Law of sines}
\\ \quad \\
\cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\\\\
-------------------------------\\\\

\begin{cases}
\measuredangle A=0.1\\
\measuredangle B=1
\end{cases}\quad thus\implies \measuredangle C=\pi -A-B\implies \measuredangle C=\pi -1.1

\bf \cfrac{sin(B)}{b}=\cfrac{sin(C)}{c}\implies \cfrac{sin(1)}{b}=\cfrac{sin(\pi -1.1)}{9}
\\\\\\
\boxed{\cfrac{9sin(1)}{sin(\pi -1.1)}=b}
\\\\\\
\cfrac{sin(A)}{a}=\cfrac{sin(C)}{c}\implies \cfrac{sin(0.1)}{a}=\cfrac{sin(\pi -1.1)}{9}
\\\\\\
\boxed{\cfrac{9sin(0.1)}{sin(\pi -1.1)}=a}

make sure your calculator is in Radian mode.
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