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 - Page 61: 16

Answer

Required trigonometric functions are- $\sin A = \frac{1}{\sqrt 6}$ $\\csc A = \sqrt 6$ $\\cos A = \frac{\sqrt 5}{\sqrt 6}$ $\\sec A = \frac{\sqrt 6}{\sqrt 5}$ $\\tan A = \frac{1}{\sqrt 5}$ $\\\cot A = \sqrt 5 $

Work Step by Step

Steps to Answer- From the given diagram of triangle ABC, we will use given information and Pythagoras Theorem to solve for 'c'- $c^{2} =a^{2} + b^{2}$ ( Pythagoras Theorem) $c^{2} = 1^{2} + (\sqrt 5)^{2}$ $c^{2} = 1 + 5$ $c^{2} = 6$ therefore $ c = \sqrt 6$ Now we can write the required six T-functions of A using a = 1, $b= \sqrt5$ and c =$ \sqrt6$ $\sin A = \frac{a}{c} = \frac{1}{\sqrt 6}$ $\\csc A = \frac{c}{a} = \frac{\sqrt 6}{1} = \sqrt 6$ $\\cos A = \frac{b}{c} = \frac{\sqrt 5}{\sqrt 6}$ $\\sec A = \frac{c}{b} = \frac{\sqrt 6}{\sqrt 5}$ $\\tan A = \frac{a}{b} = \frac{1}{\sqrt 5}$ $\\\cot A = \frac{b}{a} = \frac{\sqrt 5}{1} = \sqrt 5 $
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