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Found in: Page 487

### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508

# Trigonometric RatiosSketch a triangle that has acute angle $\mathbit{\theta }$, and find the other five trigonometric ratios of $\mathbit{\theta }$$\mathbit{c}\mathbit{o}\mathbit{s}\mathbit{\theta }\mathbf{=}\frac{\mathbf{12}}{\mathbf{13}}$

The required triangle is

The other ratios are-

$\begin{array}{l}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\mathbf{=}\frac{\mathbf{5}}{\mathbf{13}}\\ \mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\theta }\mathbf{=}\frac{\mathbf{5}}{\mathbf{12}}\\ \mathbf{c}\mathbf{o}\mathbf{t}\mathbf{\theta }\mathbf{=}\frac{\mathbf{12}}{\mathbf{5}}\\ \mathbf{c}\mathbf{s}\mathbf{c}\mathbf{\theta }\mathbf{=}\frac{\mathbf{13}}{\mathbf{5}}\\ \mathbf{s}\mathbf{e}\mathbf{c}\mathbf{\theta }\mathbf{=}\frac{\mathbf{13}}{\mathbf{12}}\end{array}$

See the step by step solution

## Step 1. Given information

A trigonometric ratio is given as $\mathrm{cos}\theta =\frac{12}{13}$.

## Step 2. Concept used

First find the unknown side of triangle. And then sketch the triangle by using the following trigonometric ratios-

$\begin{array}{l}⇒\mathrm{sin}\theta =\frac{p}{h}&\mathrm{csc}\theta =\frac{h}{p}\\ ⇒\mathrm{cos}\theta =\frac{b}{h}&\mathrm{sec}\theta =\frac{h}{b}\\ ⇒\mathrm{tan}\theta =\frac{p}{b}&\mathrm{cot}\theta =\frac{b}{p}\end{array}$

Where

$\begin{array}{l}h=hypotenuse\\ b=base\\ p=perpendicular\end{array}$

## Step 3. Calculation

The triangle by using trigonometric ratio is-

Hypotenuse is given as 13 and base is 12. Now use Pythagoras theorem for finding the third side.

$\begin{array}{l}{\left(AC\right)}^{2}={\left(AB\right)}^{2}+{\left(BC\right)}^{2}\\ {\left(13\right)}^{2}={x}^{2}+{\left(12\right)}^{2}\end{array}$

$\begin{array}{l}169={x}^{2}+144\\ {x}^{2}=169-144\\ {x}^{2}=25\\ x=5\end{array}$

Now we know all three sides. And the other five ratios are-

$\begin{array}{l}\mathrm{sin}\theta =\frac{5}{13}\\ \mathrm{tan}\theta =\frac{5}{12}\\ \mathrm{cot}\theta =\frac{12}{5}\\ \mathrm{csc}\theta =\frac{13}{5}\\ \mathrm{sec}\theta =\frac{13}{12}\end{array}$