Ambiguous case of the law of sines

Text

A triangle has $a=8$, $b=11$ and the angle $\widehat A=32^\circ$ opposite $a$. Find the possible values of the angle $\widehat B$.

Solution

By the law of sines $\dfrac{a}{\sin\widehat A}=\dfrac{b}{\sin\widehat B}$, hence $\sin\widehat B=\frac{b\sin\widehat A}{a}=\frac{11\sin 32^\circ}{8}\approx\frac{11\cdot 0.5299}{8}\approx 0.7286.$ Since $\sin\widehat B<1$ and $b>a$, there are two compatible angles: $\widehat B_1=\arcsin(0.7286)\approx 46.79^\circ,\qquad \widehat B_2=180^\circ-46.79^\circ\approx 133.21^\circ.$ Both give $\widehat A+\widehat B<180^\circ$, so both triangles exist. $\boxed{\,\widehat B_1\approx 46.79^\circ\quad\lor\quad \widehat B_2\approx 133.21^\circ\,}$