## Trigonometry 7th Edition

Published by Cengage Learning

# Chapter 2 - Section 2.1 - Definition II: Right Triangle Trigonometry - 2.1 Problem Set: 7

#### Answer

$\sin A = \frac{2}{\sqrt 5}$ $\cos A = \frac{1}{\sqrt 5}$ $\tan A = 2$ $\csc A = \frac{\sqrt 5}{2}$ $\sec A$ = $\sqrt 5$ $\cot A = \frac{1}{2}$

#### 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) $c^{2} = (2)^{2} + (1)^{2}$ $c^{2} = 4 +1$ $c^{2} = 5$ therefore $c = \sqrt 5$ Now we can write the required T-functions of A using $a=2$ , b = 1 and $c = \sqrt 5$ $\sin A = \frac{a}{c} = \frac{2}{\sqrt 5}$ $\cos A = \frac{b}{c} = \frac{1}{\sqrt 5}$ $\tan A = \frac{a}{b} =\frac{2}{1}$ = 2 $\csc A = \frac{c}{a} = \frac{\sqrt 5}{2}$ $\sec A = \frac{c}{b} = \frac{\sqrt 5}{1}$ = $\sqrt 5$ $\cot A = \frac{b}{a} = \frac{1}{2}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.