Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 10 - Geometry - 10.6 Right Triangle Trigonometry - Exercise Set 10.6 - Page 665: 3


$sinA=\frac{BC}{AB}=\frac{20}{29}$ $cosA=\frac{AC}{AB}=\frac{21}{29}$ $tanA=\frac{BC}{AC}=\frac{20}{21}$

Work Step by Step

According to Pythagorean theorem $(AB)^2=(AC)^2+(BC)^2$ This can be rearranged as: $(BC)^2=(AB)^2-(AC)^2$ We plug in the known values to obtain: $(BC)^2=(29)^2-(21)^2=(20)^2$ $\implies BC=20$ Now, we can find the trigonometric functions as $sinA=\frac{BC}{AB}=\frac{20}{29}$ $cosA=\frac{AC}{AB}=\frac{21}{29}$ $tanA=\frac{BC}{AC}=\frac{20}{21}$
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