Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 2 - Section 2.1 - Definition II: Right Triangle Trigonometry - 2.1 Problem Set: 8

Answer

$\sin A = \frac{3}{\sqrt 13}$ $\cos A = \frac{2}{\sqrt 13}$ $\tan A =\frac{3}{2}$ $\csc A = \frac{\sqrt 13}{3}$ $\sec A = \frac{\sqrt 13}{2}$ $\cot A = \frac{2}{3}$

Work Step by Step

Steps to Answer- We will use given data about triangle ABC & Pythagoras Theorem to solve for 'c'- We know that - $c^{2} =a^{2} + b^{2}$ ( Pythagoras Theorem) $c^{2} = (3)^{2} + (2)^{2}$ $c^{2} = 9 +4$ $c^{2} =13$ therefore $ c = \sqrt 13$ Now we can write the required T-functions of A using $a=3$ , b =2 and $ c = \sqrt 13$ $\sin A = \frac{a}{c} = \frac{3}{\sqrt 13}$ $\cos A = \frac{b}{c} = \frac{2}{\sqrt 13}$ $\tan A = \frac{a}{b} =\frac{3}{2}$ $\csc A = \frac{c}{a} = \frac{\sqrt 13}{3}$ $\sec A = \frac{c}{b} = \frac{\sqrt 13}{2}$ $\cot A = \frac{b}{a} = \frac{2}{3}$
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