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Q24.

Expert-verifiedFound in: Page 487

Book edition
7th Edition

Author(s)
James Stewart, Lothar Redlin, Saleem Watson

Pages
948 pages

ISBN
9781337067508

**Trigonometric Ratios**

**Sketch a triangle that has acute angle $\mathit{\theta}$, and find the other** **five trigonometric ratios of $\mathit{\theta}$**

**$\mathit{c}\mathit{o}\mathit{s}\mathit{\theta}\mathbf{=}\frac{\mathbf{12}}{\mathbf{13}}$**

The required triangle is

The other ratios are-

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

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

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

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

Where

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

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-

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

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