Answer:
Option D. sine theta equals plus or minus square root of thirty-five over six, tangent theta equals plus or minus square root of thirty five
Step-by-step explanation:
we have that
[tex]cos(\theta)=\frac{1}{6}[/tex]
If the cosine is positive, then the angle theta lie on the first or fourth Quadrant
therefore
The sine of angle theta could be positive (I Quadrant) or negative (IV Quadrant) and the tangent of angle theta could be positive (I Quadrant) or negative (IV Quadrant)
step 1
Find [tex]sin(\theta)[/tex]
Remember that
[tex]sin^{2} (\theta)+cos^{2} (\theta)=1[/tex]
we have
[tex]cos(\theta)=\frac{1}{6}[/tex]
substitute
[tex]sin^{2} (\theta)+(\frac{1}{6})^{2}=1[/tex]
[tex]sin^{2} (\theta)+\frac{1}{36}=1[/tex]
[tex]sin^{2} (\theta)=1-\frac{1}{36}[/tex]
[tex]sin^{2} (\theta)=\frac{35}{36}[/tex]
[tex]sin(\theta)=(+/-)\frac{\sqrt{35}}{6}[/tex]
so
sine theta equals plus or minus square root of thirty-five over six
step 2
Find [tex]tan(\theta)[/tex]
Remember that
[tex]tan(\theta)=\frac{sin(\theta)}{cos(\theta)}[/tex]
we have
[tex]sin(\theta)=(+/-)\frac{\sqrt{35}}{6}[/tex]
[tex]cos(\theta)=\frac{1}{6}[/tex]
substitute
[tex]tan(\theta)=(+/-)\sqrt{35}[/tex]
tangent theta equals plus or minus square root of thirty five
