Question

Trigonometric Ratios

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

$\mathit{c}\mathit{s}\mathit{c}\mathit{\theta }\mathbf{=}\frac{\mathbf{11}}{\mathbf{6}}$

Short answer

The required triangle is

The other ratios are-

$\begin{array}{l}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\mathbf{=}\frac{\mathbf{6}}{\mathbf{11}}\\ \mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }\mathbf{=}\frac{\sqrt{\mathbf{85}}}{\mathbf{11}}\\ \mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\theta }\mathbf{=}\frac{\mathbf{6}}{\sqrt{\mathbf{35}}}\\ \mathbf{s}\mathbf{e}\mathbf{c}\mathbf{\theta }\mathbf{=}\frac{\mathbf{11}}{\sqrt{\mathbf{85}}}\\ \mathbf{c}\mathbf{o}\mathbf{t}\mathbf{\theta }\mathbf{=}\frac{\sqrt{\mathbf{85}}}{\mathbf{6}}\end{array}$

Step-by-step solution

Step 1. Given information

A trigonometric ratio is given as $\csc \theta =\frac{11}{6}$.

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 ratios is-