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Q21.

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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 RatiosExpress $\mathbit{x}$ and $\mathbit{y}$ in terms of trigonometric ratios of $\mathbit{\theta }$

The values are $\mathbit{x}\mathbf{=}\frac{\mathbf{28}}{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }}$ and $\mathbit{y}\mathbf{=}\frac{\mathbf{28}}{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}$

See the step by step solution

## Step 1. Given information

A triangle is given with a side of 28.

## Step 2. Concept used

Only one side is given in the question. And for other sides, use trigonometric ratios as-

$\begin{array}{l}⇒\mathrm{sin}\theta =\frac{perpendicular}{hypotenuse}=\frac{P}{H}\\ ⇒\mathrm{cos}\theta =\frac{base}{hypotenuse}=\frac{B}{H}\\ ⇒\mathrm{tan}\theta =\frac{perpendicular}{base}=\frac{P}{B}\end{array}$

## Step 3. Calculation

From given figure,

$\begin{array}{l}⇒\mathrm{cos}\theta =\frac{B}{H}\\ ⇒\mathrm{cos}\theta =\frac{x}{28}\\ x=\frac{28}{\mathrm{cos}\theta }\end{array}$

$\begin{array}{l}⇒\mathrm{sin}\theta =\frac{P}{H}\\ ⇒\mathrm{sin}\theta =\frac{y}{28}\\ y=\frac{28}{\mathrm{sin}\theta }\end{array}$