Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 2 - Review Exercises - Page 91: 63

Answer

Possible solution (wording may differ): "Find the leg opposite to the angle $25^{o}$ in a right triangle."

Work Step by Step

Let an angle in a right triangle be $25^{o}$ Observe TOA in SOHCAHTOA, Let x be the opposite leg, and y the adjacent. Then, in that triangle $\displaystyle \tan 25^{o}=\frac{x}{y}$ when we multiply this with y, we get $x=y\cdot\tan 25^{o}$, which is of the form given in the problem... (y=3) So $x=3\tan 25^{o}$ would be a solution to a problem: "Find the leg opposite to the angle $25^{o}$ in a right triangle."
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