Answer:
According to the law of sines, \dfrac{AB}{\sin(\angle C)}=\dfrac{AC}{\sin(\angle B)}
sin(∠C)
AB
=
sin(∠B)
AC
start fraction, A, B, divided by, sine, left parenthesis, angle, C, right parenthesis, end fraction, equals, start fraction, A, C, divided by, sine, left parenthesis, angle, B, right parenthesis, end fraction. Now we can plug the values and solve:
\begin{aligned} \dfrac{AB}{\sin(\angle C)}&=\dfrac{AC}{\sin(\angle B)} \\\\ \dfrac{5}{\sin(33^\circ)}&=\dfrac{AC}{\sin(67^\circ)}\\\\ \dfrac{5\sin(67^\circ)}{\sin(33^\circ)}&=AC \\\\ 8.45&\approx AC \end{aligned}
sin(∠C)
AB
sin(33
∘
)
5
sin(33
∘
)
5sin(67
∘
)
8.45
=
sin(∠B)
AC
=
sin(67
∘
)
AC
=AC
≈AC