Trigonometry (10th Edition)

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

Chapter 2 - Quiz (Sections 2.1-2.3) - Page 67: 1

Answer

$\sin$ A = $\frac{3}{5}$ $\cos$ A = $\frac{4}{5}$ $\tan$ A = $\frac{3}{4}$ $\csc$ A = $\frac{5}{3}$ $\sec$ A = $\frac{5}{4}$ $\cot$ A = $\frac{4}{3}$

Work Step by Step

A good way to remember Trig. Functions is through the following: SOH CAH TOA In the case for this problem, angle A will be examined. $\sin$ A = $\frac{Opposite A}{Hypotenuse}$ $\sin$ A = $\frac{24}{40}$ = $\frac{3}{5}$ $\cos$ A = $\frac{Adjacent A}{Hypotenuse}$ $\cos$ A = $\frac{32}{40}$ = $\frac{4}{5}$ $\tan$ A = $\frac{Opposite A}{Adjacent A}$ $\tan$ A = $\frac{24}{32}$ = $\frac{3}{4}$ Now, the other three Trig. Functions are the inverse of the three fundamental Trig. Functions above, and are as follows: $\csc$ A = $\frac{Hypotenuse}{Opposite A}$ $\csc$ A = $\frac{40}{24}$ = $\frac{5}{3}$ $\sec$ A = $\frac{Hypotenuse}{Adjacent A}$ $\sec$ A = $\frac{40}{32}$ = $\frac{5}{4}$ $\cot$ A = $\frac{Adjacent A}{Opposite A}$ $\cot$ A = $\frac{32}{24}$ = $\frac{4}{3}$
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