## Precalculus (6th Edition) Blitzer

Use the ratio $\frac{a}{\sin A}$ or $\frac{9.3}{\sin 18{}^\circ }$. Now, use the law of sines to find B. \begin{align} & \frac{b}{\sin B}=\frac{a}{\sin A} \\ & \frac{9.3}{\sin 18{}^\circ }=\frac{41}{\sin B} \\ & \sin B=\frac{41\sin 18{}^\circ }{9.3} \\ & B\approx 1.36 \end{align} As the sine ratio can never exceed 1, there is no angle B for which $\sin B=1.36$. There is no triangle with the provided measurements.