Question

Trigonometric Ratios

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

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

Short answer

The required triangle is

The other ratios are-

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

Step-by-step solution

Step 1. Given information

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

Step 2. Concept used

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

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

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-

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