Ambiguous case of the sine rule

Statement

In a triangle $a=8$, $b=11$ and the angle $\widehat{A}=40^\circ$ (opposite $a$) are known. Find the possible values of $\widehat{B}$ and check that both give a valid triangle.

Solution

Sine rule: $\sin\widehat{B}=\dfrac{b\sin\widehat{A}}{a}=\dfrac{11\sin 40^\circ}{8}=\dfrac{11\cdot 0{.}64279}{8}\approx 0{.}8838$.

Hence $\widehat{B}\approx 62{.}10^\circ$ or $\widehat{B}\approx 180^\circ-62{.}10^\circ=117{.}90^\circ$.

Check: $40^\circ+62{.}10^\circ=102{.}10^\circ<180^\circ$ and $40^\circ+117{.}90^\circ=157{.}90^\circ<180^\circ$: both are valid triangles (ambiguous case).

$\boxed{\widehat{B}\approx 62{.}10^\circ\ \text{or}\ 117{.}90^\circ}$