## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Section 4.3 - Right Triangle Trigonometry - Exercise Set: 6

#### Answer

sin θ = $\frac{9}{41}$ cos θ = $\frac{40}{41}$ tan θ =$\frac{9}{40}$ csc θ = $\frac{41}{9}$ sec θ = $\frac{41}{40}$ cot θ = $\frac{40}{9}$

#### Work Step by Step

$a^{2}$ + $b^{y2}$ = $c^{2}$ given b= 40, c= 41 $a^{2}$ + $40^{2}$ = $41^{2}$ $a^{2}$ = 1681- 1600 $a^{2}$ = 81 a = $\sqrt {81}$ = 9 sin θ = $\frac{opposite}{hypotenuse}$ = $\frac{9}{41}$ cos θ = $\frac{adjacent}{hypotenuse}$ = $\frac{40}{41}$ tan θ = $\frac{opposite}{adjacent}$ = $\frac{9}{40}$ csc θ = $\frac{1}{sin θ}$ = $\frac{41}{9}$ sec θ = $\frac{1}{cos θ}$ = $\frac{41}{40}$ cot θ = $\frac{1}{tan θ}$ = $\frac{40}{9}$

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