Question

Solve the triangle A=48 degrees, a= 32, b=27

Answer 1

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

$\bf \cfrac{sin(48^o)}{32}=\cfrac{sin(B)}{27}\implies\cfrac{27sin(48^o)}{32}=sin(B) \\\\\\ sin^{-1}\left[ \cfrac{27sin(48^o)}{32} \right]=sin^{-1}[sin(B)]\implies sin^{-1}\left[ \cfrac{27sin(48^o)}{32} \right]=\measuredangle B \\\\\\ 38.83\approx \measuredangle B \\\\\\ \textit{now, we know angles A and B, angle is C is just 180-A-B} \\\\\\ \measuredangle C = 180-48-38.83\implies C\approx 93.17^o$

so, now we know the ∡C, so, let's use the law of sines to get side "c"

$\bf \cfrac{sin(48^o)}{32}=\cfrac{sin(93.17^o)}{c}\implies c=\cfrac{32sin(93.17^o)}{sin(48^o)}$

and surely you know how much that is