Answer:
Part 1) [tex]sin(\theta)=-\frac{5\sqrt{61}}{61}[/tex] or [tex]sin(\theta)=-\frac{5}{\sqrt{61}}[/tex]
Part 2) [tex]csc(\theta)=-\frac{\sqrt{61}}{5}[/tex]
Part 3) [tex]cot(\theta)=\frac{6}{5}[/tex]
Step-by-step explanation:
we know that
The point (-6,-5) lies on the III Quadrant
so
sin(theta) is negative
csc(theta) is negative
cot(theta) is positive
Part 1) Find the sine of angle theta
we know that
The function sine is equal to divide the opposite side to the angle theta by the hypotenuse
[tex]sin(\theta)=\frac{y}{r}[/tex]
we have
[tex]x=6\ units,y=5\ units[/tex]
Applying the Pythagoras Theorem
[tex]r^{2}=x^{2}+y^{2}[/tex]
substitute
[tex]r^{2}=6^{2}+5^{2}[/tex]
[tex]r^{2}=61[/tex]
[tex]r=\sqrt{61}\ units[/tex] -----> the hypotenuse
substitute
[tex]sin(\theta)=\frac{5}{\sqrt{61}}[/tex]
Simplify
[tex]sin(\theta)=\frac{5\sqrt{61}}{61}[/tex]
Remember that the sin(theta) is negative
so
[tex]sin(\theta)=-\frac{5\sqrt{61}}{61}[/tex]
Part 2) Find the cosecant of angle theta
we know that
The function cosecant is equal to divide the hypotenuse by the opposite side to the angle theta
[tex]csc(\theta)=1/sin(\theta)=\frac{r}{y}[/tex]
we have
[tex]sin(\theta)=-\frac{5}{\sqrt{61}}[/tex]
so
[tex]csc(\theta)=-\frac{\sqrt{61}}{5}[/tex]
Part 3) Find the cotangent of angle theta
we know that
The function cotangent is equal to divide the adjacent side to the angle theta by the opposite side to the angle theta
[tex]cot(\theta)=\frac{x}{y}[/tex]
we have
[tex]x=6\ units,y=5\ units[/tex]
substitute
[tex]cot(\theta)=\frac{6}{5}[/tex]
