Chapter 12.2, Problem 4OE

To determine

Find the values of the six trigonometric functions of $\theta$ .

Answer to Problem 4OE

  $\sin \theta =\frac{5}{\sqrt{61}}$

  $\cos \theta =\frac{6}{\sqrt{61}}$

  $\tan \theta =\frac{5}{6}$

  $\csc \theta =\frac{\sqrt{61}}{5}$

  $\sec \theta =\frac{\sqrt{61}}{6}$

  $\cot \theta =\frac{6}{5}$

Explanation of Solution

Given:

  

Formula Used:

  $\sin \theta =\frac{\text{length of side opposite to }\theta }{\text{length of the hypotenuse}}$

  $\cos \theta =\frac{\text{length of side adjacent to }\theta }{\text{length of the hypotenuse}}$

  $\tan \theta =\frac{\text{length of side opposite to }\theta }{\text{length of side adjacent to }\theta }$

  $\csc \theta =\frac{1}{\sin \theta }$

  $\sec \theta =\frac{1}{\cos \theta }$

  $\cot \theta =\frac{1}{\tan \theta }$

Calculation:

In the triangle ,

x = length of side adjacent to $\theta$ = 6 units

y = length of side opposite to $\theta$= 5 units

r = length of the hypotenuse = $\sqrt{x^2+y^2}=\sqrt{36+25}=\sqrt{61}$

So,

  $\sin \theta =\frac{\text{length of side opposite to }\theta }{\text{length of the hypotenuse}}=\frac{y}{r}=\frac{5}{\sqrt{61}}$

  $\cos \theta =\frac{\text{length of side adjacent to }\theta }{\text{length of the hypotenuse}}=\frac{x}{r}=\frac{6}{\sqrt{61}}$

  $\tan \theta =\frac{\text{length of side opposite to }\theta }{\text{length of side adjacent to }\theta }=\frac{y}{x}=\frac{5}{6}$

  $\csc \theta =\frac{1}{\sin \theta }=\frac{\sqrt{61}}{5}$

  $\sec \theta =\frac{1}{\cos \theta }=\frac{\sqrt{61}}{6}$

  $\cot \theta =\frac{1}{\tan \theta }=\frac{6}{5}$