## Precalculus (6th Edition) Blitzer

The required values are $A=82{}^\circ,B=41{}^\circ,C=57{}^\circ,c\approx 255.7$.
Using the law of sines we will obtain: \begin{align} & \frac{a}{\sin A}=\frac{b}{\sin B} \\ & \frac{300}{\sin 2\theta }=\frac{200}{\sin \theta } \\ & 200\sin 2\theta =300\sin \theta \end{align} Using the identity: $\sin 2\theta =2\sin \theta \cos \theta$ $400\sin \theta \cos \theta =300\sin \theta$ Now, \begin{align} & \cos \theta =\frac{300\sin \theta }{400\sin \theta } \\ & \cos \theta =\frac{3}{4} \\ & \theta \approx 41{}^\circ \\ & 2\theta \approx 82{}^\circ \end{align} Now, we will find C, by using the property that the sum of all angles is $180{}^\circ$ So, \begin{align} & A+B+C=180{}^\circ \\ & 82{}^\circ +41{}^\circ +C=180{}^\circ \\ & C=180{}^\circ -123{}^\circ \\ & C=57{}^\circ \end{align} Using the law of sines, we will find c. \begin{align} & \frac{a}{\sin A}=\frac{c}{\sin C} \\ & \frac{300}{\sin 82{}^\circ }=\frac{c}{\sin 57{}^\circ } \\ & c=\frac{300\sin 57{}^\circ }{\sin 82{}^\circ } \\ & c\approx 255.7 \end{align} Hence, the required values are $A=82{}^\circ,B=41{}^\circ,C=57{}^\circ,c\approx 255.7$.