Chapter 12.1, Problem 16PPS

To determine

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

Answer to Problem 16PPS

The all the values of trigonometric functions are $\sin \theta =\frac{6}{\sqrt{117}},\cos \theta =\frac{9}{\sqrt{117}},\tan \theta =\frac{6}{9},\csc \theta =\frac{\sqrt{117}}{6},\sec \theta =\frac{\sqrt{117}}{9},\cot \theta =\frac{9}{6}$ .

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 pythagoras 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 $b=9,p=6$ .

  $\begin{array}{l}{h}^2=p^2+b^2\\ h^2=6^2+9^2\\ h^2=36+81\\ h^2=117\\ h=\sqrt{117}\end{array}$

Now, use the trigonometric ratios:

  $\sin \theta =\frac{6}{\sqrt{117}},\cos \theta =\frac{9}{\sqrt{117}},\tan \theta =\frac{6}{9},\csc \theta =\frac{\sqrt{117}}{6},\sec \theta =\frac{\sqrt{117}}{9},\cot \theta =\frac{9}{6}$

Conclusion:

The all the values are $\sin \theta =\frac{6}{\sqrt{117}},\cos \theta =\frac{9}{\sqrt{117}},\tan \theta =\frac{6}{9},\csc \theta =\frac{\sqrt{117}}{6},\sec \theta =\frac{\sqrt{117}}{9},\cot \theta =\frac{9}{6}$ .