## Precalculus (6th Edition) Blitzer

The platform is about $3671.8\text{ yards}$ from one end of the beach and $\text{3576}\text{.4 yards}$ from the other end.
Using the figure, we will find C: \begin{align} & C=180{}^\circ -A-B \\ & =180{}^\circ -85{}^\circ -76{}^\circ \\ & =19{}^\circ \end{align} Now, the ratio $\frac{c}{\sin C}$, or $\frac{1200}{\sin 19{}^\circ }$, is known. Using the law of sines we will find a and b: \begin{align} & \frac{a}{\sin A}=\frac{c}{\sin C} \\ & \frac{a}{\sin 85{}^\circ }=\frac{1200}{\sin 19{}^\circ } \\ & a=\frac{1200\sin 85{}^\circ }{\sin 19{}^\circ } \\ & a\approx 3671.8 \end{align} Now, we will evaluate b using the law of sines as below: \begin{align} & \frac{b}{\sin B}=\frac{c}{\sin C} \\ & \frac{b}{\sin 76{}^\circ }=\frac{1200}{\sin 19{}^\circ } \\ & b=\frac{1200\sin 76{}^\circ }{\sin 19{}^\circ } \\ & b\approx 3576.4 \end{align} The platform is about $3671.8\text{ yards}$ from one end of the beach and $\text{3576}\text{.4 yards}$ from the other end.