$\bf \textit{Law of sines}
\\ \quad \\
\cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\\\\
-------------------------------\\\\$

$\bf \cfrac{sin(88^o)}{39.9}=\cfrac{sin(A)}{31}\implies \cfrac{31\cdot sin(88^o)}{39.9}=sin(A)
\\\\\\
sin^{-1}\left[ \cfrac{31\cdot sin(88^o)}{39.9} \right]=sin^{-1}\left[ sin(A) \right]
\\\\\\
sin^{-1}\left[ \cfrac{31\cdot sin(88^o)}{39.9} \right]=\measuredangle A
\\\\\\
50.93841^o\approx \measuredangle A\qquad thus\qquad \measuredangle C\approx 41.0619^o$

3 0

3 years ago

-8x + 5x - 9 = 6<br> Can you help me?

Ann [662]

Hope this helps sorry my writing is messy

3 0

2 years ago

Help plsssssssssssssssss​

Help plsssssssssssssssss