Chapter 4, Problem 69RE

To determine

To Find:The exact values of the six trigonometric functions of the angle $\theta$ .

Answer to Problem 69RE

The six trigonometric functions of $\theta$ are

  $\sin \theta =\frac{15}{3\sqrt{41}}$

  $\cos \theta =\frac{12}{3\sqrt{41}}$

  $\tan \theta =\frac{15}{12}$

  $\cos ec\theta =\frac{3\sqrt{41}}{15}$

  $\sec \theta =\frac{3\sqrt{41}}{12}$

  $\cot \theta =\frac{12}{15}$

Explanation of Solution

Given information: Right angle triangle

  

Concept used:

Let $\theta$ be an acute angle of a right triangle. The six trigonometric functions of the angle $\theta$ are defined as follows

  $\sin \theta =\frac{opp}{hyp}$

  $\cos \theta =\frac{adj}{hyp}$

  $\tan \theta =\frac{opp}{adj}$

  $\cos ec\theta =\frac{hyp}{opp}$

  $\sec \theta =\frac{hyp}{adj}$

  $\cot \theta =\frac{adj}{opp}$

The abbreviations opp,adj, and hyp represent the length of the three sides of a right triangle

  $opp=$ the length of the side opposite $\theta$

  $adj=$ the length of the side adjacent to $\theta$

  $hyp=$ the length of the hypotenuse

Calculation:

Use the Pythagorean theorem

  ${\left(hyp\right)}^2={\left(adj\right)}^2+{\left(opp\right)}^2$

It follows that

  $hyp=\sqrt{{(12)}^2+{(15)}^2}=3\sqrt{41}$

So, the six trigonometric functions of $\theta$ are

  $\begin{array}{l}\sin \theta =\frac{15}{3\sqrt{41}}\\ \cos \theta =\frac{12}{3\sqrt{41}}\\ \tan \theta =\frac{15}{12}\\ \cos ec\theta =\frac{3\sqrt{41}}{15}\\ \sec \theta =\frac{3\sqrt{41}}{12}\\ \cot \theta =\frac{12}{15}\end{array}$