Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 13, Trigonometric Ratios and Functions - 13.5 Apply the Law of Sines - 13.5 Exercises - Problem Solving - Page 887: 43


See below.

Work Step by Step

The third angle: $180-40-45=95^\circ$. Let the two distances be $d_1,d_2$. Using the law of sines, $\frac{d_1}{\sin45^\circ}=\frac{300}{\sin95^\circ}\\d_1=\frac{300\sin45^\circ}{\sin95^\circ}\approx212.9$ $\frac{d_2}{\sin40^\circ}=\frac{300}{\sin95^\circ}\\d_2=\frac{300\sin40^\circ}{\sin95^\circ}\approx193.6$
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