Chapter 12.3, Problem 33WE

To determine

Write the given as a function of an acute angle.

Answer to Problem 33WE

  $\tan 85.33°$

Explanation of Solution

Given:

  $\tan 265°20\text{'}$

Formula Used:

If $\theta$ is not a quadrantal angle, there is a unique acute angle $\alpha$ corresponding to $\theta$ , formed by the terminal side of $\theta$ and the positive or negative x- axis.When $\theta$ is in standard position , $\alpha$ is called the reference angle of $\theta$ .

Function value	Quadrant of $\theta$ Is I	Quadrant of $\theta$ Is II	Quadrant of $\theta$ Is III	Quadrant of $\theta$ Is IV
$\begin{array}{l}\sin \theta \\ \csc \theta \end{array}$	+	+	-	-
$\begin{array}{l}\cos \theta \\ \sec \theta \end{array}$	+	-	-	+
$\begin{array}{l}\tan \theta \\ \cot \theta \end{array}$	+	-	+	-

Calculation:

Since, $\theta =265°20\text{'}$ ,first convert minutes into degree:

  $\theta =265°20\text{'}=265°+\left(\frac{20}{60}\right)°=265°+0.33°=265.33°$ , which is in Quadrant III, where tan is positive.

So, the reference angle is :

  $\alpha =\theta \text{'}-180°=265.33°-180°=85.33°$

So, the given angle as a function of acute angle is :

  $\tan 85.33°$