Question

Two triangles can be formed with the given information. Use the Law of Sines to solve the triangles.

Answer 1

check the picture below.

$\bf sin^{-1}\left( \cfrac{14\cdot sin(55^o)}{12} \right)=\measuredangle B\implies sin^{-1}(0.955677)\approx \measuredangle B\\\\\\17.122516879^o\approx \measuredangle B\qquad \qquad \qquad \qquad \qquad \measuredangle C\approx 162.877483^o\\\\\\\stackrel{\textit{the second triangle for this \underline{ambiguous case} will be when }}{17.122516879^o\approx \measuredangle C\qquad \qquad \qquad \qquad \qquad \measuredangle B\approx 162.877483^o}$

recall the second case is when B is 180° - B, namely the supplementary angle of 17.122516879°, or 180° - 17.122516879°.

now, as you see in the picture, the length c is the same on both triangles, on the first one is sticking out, and on the second triangle is sticking in, as you see there in red.

$\bf \stackrel{using~~C\approx 162.88^o}{\cfrac{sin(A)}{12}=\cfrac{sin(C)}{c}}\implies \cfrac{sin(55^o)}{12}=\cfrac{sin(162.88^o)}{c}\\\\\\c=\cfrac{12\cdot sin(162.88^o) }{sin(55^o)}\implies c\approx 4.312371$