Chapter 12.2, Problem 60SR

To determine

Calculate the values of six trigonometric functions for angle $\theta$ .

Answer to Problem 60SR

The all the values of trigonometric functions are $\sin \theta =\frac{11}{\sqrt{290}},\cos \theta =\frac{13}{\sqrt{290}},\tan \theta =\frac{11}{13},\csc \theta =\frac{\sqrt{290}}{11},\sec \theta =\frac{\sqrt{290}}{13},\cot \theta =\frac{13}{11}$ .

Explanation of Solution

Given:

It is given in the question that the figure is given below in which b and p are given:

  

Concept Used:

In this, use the concept of pythgoras theorem and trigonometric ratios and the formula will be $h^2=p^2+b^2$ , $\sin \theta =\frac{p}{h},\cos \theta =\frac{b}{h},\tan \theta =\frac{p}{b},\csc \theta =\frac{h}{p},\sec \theta =\frac{h}{b},\cot \theta =\frac{b}{p}$ .

Calculation: Here, use Pythagoras theorem where $b=13,p=11$ .

  $\begin{array}{l}{h}^2=p^2+b^2\\ h^2={11}^2+{13}^2\\ h^2=121+169\\ h^2=290\\ h=\sqrt{290}\end{array}$

Now, use the trigonometric ratios:

  $\sin \theta =\frac{11}{\sqrt{290}},\cos \theta =\frac{13}{\sqrt{290}},\tan \theta =\frac{11}{13},\csc \theta =\frac{\sqrt{290}}{11},\sec \theta =\frac{\sqrt{290}}{13},\cot \theta =\frac{13}{11}$

Conclusion:

The all the values are $\sin \theta =\frac{11}{\sqrt{290}},\cos \theta =\frac{13}{\sqrt{290}},\tan \theta =\frac{11}{13},\csc \theta =\frac{\sqrt{290}}{11},\sec \theta =\frac{\sqrt{290}}{13},\cot \theta =\frac{13}{11}$ .