Answer:
Part 4) [tex]r=84\ units[/tex]
Part 9) [tex]sin(\theta)=-\frac{\sqrt{5}}{3}[/tex]
Part 10) [tex]sin(\theta)=-\frac{9\sqrt{202}}{202}[/tex]
Step-by-step explanation:
Part 4) A circle has an arc of length 56pi that is intercepted by a central angle of 120 degrees. What is the radius of the circle?
we know that
The circumference of a circle subtends a central angle of 360 degrees
The circumference is equal to
[tex]C=2\pi r[/tex]
using proportion
[tex]\frac{2\pi r}{360^o}=\frac{56\pi}{120^o}[/tex]
simplify
[tex]\frac{r}{180^o}=\frac{56}{120^o}[/tex]
solve for r
[tex]r=\frac{56}{120^o}(180^o)[/tex]
[tex]r=84\ units[/tex]
Part 9) Given cos(∅)=-2/3 and ∅ lies in Quadrant III. Find the exact value of sin(∅) in simplified form
Remember the trigonometric identity
[tex]cos^2(\theta)+sin^2(\theta)=1[/tex]
we have
[tex]cos(\theta)=-\frac{2}{3}[/tex]
substitute the given value
[tex](-\frac{2}{3})^2+sin^2(\theta)=1[/tex]
[tex]\frac{4}{9}+sin^2(\theta)=1[/tex]
[tex]sin^2(\theta)=1-\frac{4}{9}[/tex]
[tex]sin^2(\theta)=\frac{5}{9}[/tex]
square root both sides
[tex]sin(\theta)=\pm\frac{\sqrt{5}}{3}[/tex]
we know that
If ∅ lies in Quadrant III
then
The value of sin(∅) is negative
[tex]sin(\theta)=-\frac{\sqrt{5}}{3}[/tex]
Part 10) The terminal side of ∅ passes through the point (11,-9). What is the exact value of sin(∅) in simplified form?
see the attached figure to better understand the problem
In the right triangle ABC of the figure
[tex]sin(\theta)=\frac{BC}{AC}[/tex]
Find the length side AC applying the Pythagorean Theorem
[tex]AC^2=AB^2+BC^2[/tex]
substitute the given values
[tex]AC^2=11^2+9^2[/tex]
[tex]AC^2=202[/tex]
[tex]AC=\sqrt{202}\ units[/tex]
so
[tex]sin(\theta)=\frac{9}{\sqrt{202}}[/tex]
simplify
[tex]sin(\theta)=\frac{9\sqrt{202}}{202}[/tex]
Remember that
The point (11,-9) lies in Quadrant IV
then
The value of sin(∅) is negative
therefore
[tex]sin(\theta)=-\frac{9\sqrt{202}}{202}[/tex]
