To solve for the values of [tex]\(\sin \theta\)[/tex] and [tex]\(\tan \theta\)[/tex] given that [tex]\(\cos \theta = \frac{39}{89}\)[/tex] and the angle [tex]\(\theta\)[/tex] is in the fourth quadrant, we use trigonometric identities and the given trigonometric functions.
### Part A: Finding [tex]\(\sin \theta\)[/tex]
We know from the Pythagorean identity that:
[tex]\[
\cos^2 \theta + \sin^2 \theta = 1
\][/tex]
Given [tex]\(\cos \theta = \frac{39}{89}\)[/tex], we first find [tex]\(\cos^2 \theta\)[/tex]:
[tex]\[
\cos^2 \theta = \left(\frac{39}{89}\right)^2 = \frac{1521}{7921}
\][/tex]
Next, we use the Pythagorean identity to solve for [tex]\(\sin^2 \theta\)[/tex]:
[tex]\[
\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{1521}{7921} = \frac{7921 - 1521}{7921} = \frac{6400}{7921}
\][/tex]
Then, taking the square root to find [tex]\(\sin \theta\)[/tex], we get:
[tex]\[
\sin \theta = \pm \sqrt{\frac{6400}{7921}} = \pm \frac{80}{89}
\][/tex]
Since [tex]\(\theta\)[/tex] is in the fourth quadrant, [tex]\(\sin \theta\)[/tex] is negative:
[tex]\[
\sin \theta = -\frac{80}{89}
\][/tex]
Thus, the value of [tex]\(\sin \theta\)[/tex] is:
[tex]\[
\boxed{-\frac{80}{89}}
\][/tex]
### Part B: Finding [tex]\(\tan \theta\)[/tex]
Now, we use the definition of tangent in terms of sine and cosine:
[tex]\[
\tan \theta = \frac{\sin \theta}{\cos \theta}
\][/tex]
We already have [tex]\(\sin \theta = -\frac{80}{89}\)[/tex] and [tex]\(\cos \theta = \frac{39}{89}\)[/tex]. Therefore:
[tex]\[
\tan \theta = \frac{-\frac{80}{89}}{\frac{39}{89}} = \frac{-80}{39}
\][/tex]
Thus, the value of [tex]\(\tan \theta\)[/tex] is:
[tex]\[
\boxed{-\frac{80}{39}}
\][/tex]
