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: 6

Answer

$\sin A = \frac{12}{13}$ $\\cos A = \frac{5}{13}$ $\\tan A = \frac{12}{5}$ $\\csc A = \frac{13}{12}$ $\\\sec A = \frac{13}{5}$ $\\\cot A = \frac{5}{12}$

Work Step by Step

We will use given data about triangle ABC & Pythagoras Theorem to solve for 'a'- We know that - $c^{2} =a^{2} + b^{2}$ ( Pythagoras Theorem) Therefore - $a^{2} =c^{2} - b^{2}$ $a^{2} = (13)^{2} - (5)^{2}$ $a^{2} = 169 - 25$ $a^{2} = 144$ therefore $ a = \sqrt (144)$ a = 12 Now we can write the required T-functions of A and B using $a=12$ , b = 5 and c = 13 $\sin A = \frac{a}{c} = \frac{12}{13}$ $\\cos A = \frac{b}{c} = \frac{5}{13}$ $\\tan A = \frac{a}{b} = \frac{12}{5}$ $\\csc A = \frac{c}{a} = \frac{13}{12}$ $\\\sec A = \frac{c}{b} = \frac{13}{5}$ $\\\cot A = \frac{b}{a} = \frac{5}{12}$
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