Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Section 6.1 - The Law of Sines - Exercise Set - Page 721: 51

Answer

The distance from the base to the top of the tower is about $\text{184}\text{.3 feet}$.

Work Step by Step

Let the distance from the base to the top of the tower -- that is, the distance between A and C -- be b. To find angle C we will use the angle sum property: $\begin{align} & C=180{}^\circ -A-B \\ & =180{}^\circ -84.7{}^\circ -50{}^\circ \\ & =45.3{}^\circ \end{align}$ Using the law of sines we will find c: $\begin{align} & \frac{c}{\sin C}=\frac{b}{\sin B} \\ & \frac{171}{\sin 45.3{}^\circ }=\frac{b}{\sin 50{}^\circ } \\ & b=\frac{171\sin 50{}^\circ }{\sin 45.3{}^\circ } \\ & \approx 184.3 \end{align}$ The distance between A and C is about $\text{184}\text{.3 feet}$.
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