## Precalculus (6th Edition) Blitzer

Let us consider A to be the point where the angle of elevation is $66{}^\circ$. Also consider B to be the point where the angle of elevation is $50{}^\circ$. Let C be the point at the top of the tree. The angle made at point C by the other two sides is given by: \begin{align} & C=180{}^\circ -A-B \\ & =180{}^\circ -66{}^\circ -50{}^\circ \\ & C=64{}^\circ \end{align} Using the law of sines, we will find a: \begin{align} & \frac{a}{\sin \,A}=\frac{c}{\sin \,C} \\ & \frac{a}{\sin \,66{}^\circ }=\frac{420}{\sin \,64{}^\circ } \\ & a=\frac{420\times \sin \,66{}^\circ }{\sin \,64{}^\circ } \\ & a=426.9 \end{align} The height of the tree denoted by h is given by \begin{align} & h=a\,\sin \,B \\ & =426.9\,\sin \,50{}^\circ \\ & h=327.0 \end{align} Therefore, the height of the tree is 327.0 feet.