Chapter 4, Problem 45CR

To determine

To Find: The value of all the trigonometric ratios in terms of $\theta$ given in the question.

Answer to Problem 45CR

  $\sin \theta$ = $\frac{5}{\sqrt{61}}$ , $\cos \theta =\frac{6}{\sqrt{61}}$ , $\tan \theta =\frac{5}{6}$ , $\cos ec\theta =\frac{\sqrt{61}}{5}$ , $sec\theta =\frac{\sqrt{61}}{6}$ , $\cot \theta =\frac{6}{5}$ .

Explanation of Solution

Given information: A right angled triangle given in the question where,

Side opposite to angle $\theta$ = 10 units

And, side adjacent to angle $\theta$ = 12 units

Formula Used: To find the hypotenuse, use Pythagoras theorem that is

  ${(Hypotenuse)}^2={(opposite)}^2+{(adjacent)}^2$

  $\sin \theta$ = $\frac{opposite}{hypotenuse}$

  $\cos \theta =\frac{adjacent}{hypotenuse}$

  $\tan \theta =\frac{opposite}{adjacent}$

  $\cos ec\theta =\frac{1}{\sin \theta }$

  $sec\theta =\frac{1}{\cos \theta }$

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

By using Pythagoras theorem

  ${(hypotenuse)}^2={12}^2+{10}^2$

  $hypotenuse=2\sqrt{61}$ units

Now , $\sin \theta$ = $\frac{opposite}{hypotenuse}$

So, $\sin \theta$ = $\frac{5}{\sqrt{61}}$

  $\cos \theta =\frac{adjacent}{hypotenuse}$

So, $\cos \theta =\frac{6}{\sqrt{61}}$

  $\tan \theta =\frac{opposite}{adjacent}$

So, $\tan \theta =\frac{5}{6}$

Similarly, $\cos ec\theta =\frac{\sqrt{61}}{5}$ , $sec\theta =\frac{\sqrt{61}}{6}$ , $\cot \theta =\frac{6}{5}$