Chapter 4, Problem 71RE

To determine

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

Answer to Problem 71RE

The six trigonometric functions of $\theta$ are

  $\sin \theta =\frac{10}{2\sqrt{61}}$

  $\cos \theta =\frac{12}{2\sqrt{61}}$

  $\tan \theta =\frac{10}{12}$

  $\cos ec\theta =\frac{2\sqrt{61}}{10}$

  $\sec \theta =\frac{2\sqrt{61}}{12}$

  $\cot \theta =\frac{12}{10}$

Explanation of Solution

Given information:

Right angle triangle is given as:

  

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

  $hyp=\sqrt{{(10)}^2+{(12)}^2}=2\sqrt{61}$

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

  $\sin \theta =\frac{10}{2\sqrt{61}}$

  $\cos \theta =\frac{12}{2\sqrt{61}}$

  $\tan \theta =\frac{10}{12}$

  $\cos ec\theta =\frac{2\sqrt{61}}{10}$

  $\sec \theta =\frac{2\sqrt{61}}{12}$

  $\cot \theta =\frac{12}{10}$