Question

9. 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?

10. The terminal side of (0) passes through the point (11,-9). What is the exact value of sin(0) in simplified form?

Answer 1

9. r=84

10.
`-(9√(202) )/(202)`

Explanation:

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Answer 2

Part 4)
`r=84\ units`

Part 9)
`sin(\theta)=-(√(5))/(3)`

Part 10)
`sin(\theta)=-(9√(202))/(202)`

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

`C=2\pi r`

using proportion

`(2\pi r)/(360^o)=(56\pi)/(120^o)`

simplify

`(r)/(180^o)=(56)/(120^o)`

solve for r

`r=(56)/(120^o)(180^o)`

`r=84\ units`

Part 9) Given cos(∅)=-2/3 and ∅ lies in Quadrant III. Find the exact value of sin(∅) in simplified form

Remember the trigonometric identity

`cos^2(\theta)+sin^2(\theta)=1`

we have

`cos(\theta)=-(2)/(3)`

substitute the given value

`(-(2)/(3))^2+sin^2(\theta)=1`

`(4)/(9)+sin^2(\theta)=1`

`sin^2(\theta)=1-(4)/(9)`

`sin^2(\theta)=(5)/(9)`

square root both sides

`sin(\theta)=\pm(√(5))/(3)`

we know that

If ∅ lies in Quadrant III

then

The value of sin(∅) is negative

`sin(\theta)=-(√(5))/(3)`

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

`sin(\theta)=(BC)/(AC)`

Find the length side AC applying the Pythagorean Theorem

`AC^2=AB^2+BC^2`

substitute the given values

`AC^2=11^2+9^2`

`AC^2=202`

`AC=√(202)\ units`

so

`sin(\theta)=(9)/(√(202))`

simplify

`sin(\theta)=(9√(202))/(202)`

Remember that

The point (11,-9) lies in Quadrant IV

then

The value of sin(∅) is negative

therefore

`sin(\theta)=-(9√(202))/(202)`