Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 7 - Applications of Trigonometric Functions - Section 7.1 Right Triangle Trigonometry ; Applications - 7.1 Assess Your Understanding - Page 540: 10


sin θ = $\frac{opposite}{hypotenuse}$ = $\frac{3}{5}$ cos θ = $\frac{adjacent}{hypotenuse}$ = $\frac{4}{5}$ tan θ = $\frac{opposite}{adjacent}$ = $\frac{3}{4}$ cot θ = $\frac{adjacent}{opposite}$ = $\frac{4}{3}$ sec θ = $\frac{hypotenuse}{adjacent}$ = $\frac{5}{4}$ csc θ = $\frac{hypotenuse}{opposite}$ = $\frac{5}{3}$

Work Step by Step

First, we need to find the value of the hypotenuse using the Pythagorean theorem, which states that $a^2 + b^2 = c^2$, where $a$ is the length of the side adjacent to the given angle, $b$ is the length of the side opposite the angle, and $c$ is the length of the hypotenuse. In this triangle, $a$ is $4$ and $b$ is $3$, so let's plug these values into the formula to find the length of the hypotenuse: $4^2 + 3^2 = c^2$ Evaluate the exponents: $16 + 9 = c^2$ Add on the left side of the equation: $c^2 = 25$ Take the square root of both sides of the equation: $c = 5$ Now that we have the lengths of all sides of the right triangle, let's plug in the values into the formulas for the trigonometric functions: sin θ = $\frac{opposite}{hypotenuse}$ = $\frac{3}{5}$ cos θ = $\frac{adjacent}{hypotenuse}$ = $\frac{4}{5}$ tan θ = $\frac{opposite}{adjacent}$ = $\frac{3}{4}$ cot θ = $\frac{adjacent}{opposite}$ = $\frac{4}{3}$ sec θ = $\frac{hypotenuse}{adjacent}$ = $\frac{5}{4}$ csc θ = $\frac{hypotenuse}{opposite}$ = $\frac{5}{3}$
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