Chapter 6, Problem 23CT

To determine

The exact value of the function, $\cos \theta$ if the point $(-5,11)$ is on the terminal side of angle $\theta$ in standard position.

Answer to Problem 23CT

Solution:

The exact value of $\cos \theta$ is $-\frac{5\sqrt{146}}{146}$ .

Explanation of Solution

Given information:

The point $(-5,11)$ is on the terminal side of angle $\theta$ in standard position.

Let, the point $P(x,y)=(-5,11)$ is on the terminal side of angle $\theta$ in standard position, that is on the circle, $x^2+y^2=r^2$ .

Thus, $x=-5$ , and $y=11$ .

Now use $x^2+y^2=r^2$ to find the value of $r$ .

Thus, ${\left(-5\right)}^2+{\left(11\right)}^2=r^2$ .

  $\Rightarrow r^2=25+121$ .

  $\Rightarrow r^2=146$

  $\Rightarrow r=\sqrt{146}$ .

Thus, here $x=-5$ , $y=11$ and $r=\sqrt{146}$ .

  $\cos \theta =\frac{x}{r}=-\frac{5}{\sqrt{146}}=-\frac{5\sqrt{146}}{146}$

Thus, the exact value of $\cos \theta$ is $-\frac{5\sqrt{146}}{146}$ .