## Precalculus (6th Edition) Blitzer

The required values are $A=96{}^\circ,B=48{}^\circ,C=36{}^\circ,c\approx 237.3$.
Using the law of sines, we will obtain: \begin{align} & \frac{a}{\sin A}=\frac{b}{\sin B} \\ & \frac{400}{\sin 2\theta }=\frac{300}{\sin \theta } \\ & 300\sin 2\theta =400\sin \theta \end{align} Using the identity: $\sin 2\theta =2\sin \theta \cos \theta$, we get So, $600\sin \theta \cos \theta =400\sin \theta$ Now, \begin{align} & \cos \theta =\frac{400\sin \theta }{600\sin \theta } \\ & \cos \theta =\frac{2}{3} \\ & \theta =48{}^\circ \\ & 2\theta =96{}^\circ \end{align} Now, to find C, we will use the rule that the sum of all angles is $180{}^\circ$ So, \begin{align} & A+B+C=180{}^\circ \\ & 96{}^\circ +48{}^\circ +C=180{}^\circ \\ & C=180{}^\circ -144{}^\circ \\ & C=36{}^\circ \end{align} Using the law of sines, we will find c. \begin{align} & \frac{a}{\sin A}=\frac{c}{\sin C} \\ & \frac{400}{\sin 96{}^\circ }=\frac{c}{\sin 36{}^\circ } \\ & c=\frac{400\sin 36{}^\circ }{\sin 96{}^\circ } \\ & c\approx 237.3 \end{align} Hence, the required values are $A=96{}^\circ,B=48{}^\circ,C=36{}^\circ,c\approx 237.3$.