Answer:
Part 1) [tex]sin(\theta)=-\frac{\sqrt{65}}{9}[/tex]
Part 2) [tex]tan(\theta)=\frac{\sqrt{65}}{4}[/tex]
Step-by-step explanation:
we have that
The cosine of angle theta is negative and the tangent of angle theta is positive
That means that the sine of angle theta is negative
step 1
Find [tex]sin(\theta)[/tex]
we know that
[tex]sin^{2}(\theta) +cos^{2}(\theta)=1[/tex]
we have
[tex]cos(\theta)=-\frac{4}{9}[/tex]
substitute
[tex]sin^{2}(\theta) +(-\frac{4}{9})^{2}=1[/tex]
[tex]sin^{2}(\theta) +\frac{16}{81}=1[/tex]
[tex]sin^{2}(\theta)=1-\frac{16}{81}[/tex]
[tex]sin^{2}(\theta)=\frac{65}{81}[/tex]
square root both sides
[tex]sin(\theta)=\pm\frac{\sqrt{65}}{9}[/tex]
Remember that
In this problem the sine of angle theta is negative
so
[tex]sin(\theta)=-\frac{\sqrt{65}}{9}[/tex]
step 2
Find [tex]tan(\theta)[/tex]
we know that
[tex]tan(\theta)=\frac{sin(\theta)}{cos(\theta)}[/tex]
we have
[tex]sin(\theta)=-\frac{\sqrt{65}}{9}[/tex]
[tex]cos(\theta)=-\frac{4}{9}[/tex]
substitute the given values
[tex]tan(\theta)=-\frac{\sqrt{65}}{9}:-\frac{4}{9}=\frac{\sqrt{65}}{4}[/tex]
