Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 8 - Application of Trigonometry - Summary Exercises on Applications of Trigonometry and Vectors - Exercises - Page 792: 10

Answer

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Work Step by Step

Assume the angle of the bottom left corner be $x$, use the Law of Sines, we have $\frac{sin(x)}{26.5}=\frac{sin(28^\circ10')}{21.2}$, thus $sin(x)=\frac{26.5sin28.17^\circ}{21.2}\approx0.59$. For $x\in(0,\pi)$, there are two solutions $x=36.16^\circ$ and $x=180^\circ-36.16^\circ=143.84^\circ$. It looks like the surveyor only considered the case of an acute angle $x$.
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