## Precalculus (6th Edition) Blitzer

The given angles and side of the triangle are provided below: $A=65{}^\circ,a=6,b=7$ Using the law of sines, we will find the angle B of the triangle. That is, \begin{align} & \frac{\sin A}{a}=\frac{\sin B}{b} \\ & \frac{\sin 65{}^\circ }{6}=\frac{\operatorname{sinB}}{7} \end{align} Therefore, \begin{align} & \operatorname{sinB}=7\times \frac{\sin 65{}^\circ }{6} \\ & \operatorname{sinB}=1.0574 \end{align} The function sine can’t exceed the value 1. Therefore, $\operatorname{sinB}=1.0574$ is not possible. Thus, there does not exist any triangle of this type with the given measurements.