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: 11

Answer

$\sin A = \frac{5}{6}$ , $\\csc A = \frac{6}{5}$ $\\cos A = \frac{\sqrt11}{6}$ , $\\sec A = \frac{6}{\sqrt 11}$ $\\tan A = \frac{5}{\sqrt 11}$ , $\\\cot A = \frac{\sqrt11}{5}$

Work Step by Step

Steps to Answer- First we will draw the diagram of triangle ABC and then use the given information and Pythagoras Theorem to solve for 'b'- We know that $c^{2} =a^{2} + b^{2}$ ( Pythagoras Theorem) therefore $b^{2} =c^{2} -a^{2}$ $b^{2} = 6^{2} - 5^{2}$ $b^{2} = 36 - 25$ $b^{2} = 11$ therefore $ b = \sqrt 11$ Now we can write the required six T-functions of A using a = 3, $b= \sqrt 7$ and c = 4 $\sin A = \frac{a}{c} = \frac{5}{6}$ $\\csc A = \frac{c}{a} = \frac{6}{5}$ $\\cos A = \frac{b}{c} = \frac{\sqrt 11}{6}$ $\\sec A = \frac{c}{b} = \frac{6}{\sqrt 11}$ $\\tan A = \frac{a}{b} = \frac{5}{\sqrt 11}$ $\\\cot A = \frac{b}{a} = \frac{\sqrt 11}{5}$
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