4. The diameter of circle C is 18 cm. Chord AD is the same length as a radius. Use this information, the diagram, and your experiences in geometry this semester to answer these questions. (Use pie = 3.14 .)
a.) What are the measures of angles BDA and DBA?
b.) What is the sine of angle DBA?
c.) Find the length of arc AB to the nearest hundredth.
d.) How does the length of arc AD compare to the length of arc AB?
e.) If EF = 1/2 BD, then what is the degree measure of arc EF?
Answer:

- Zolock

- katieb

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- Zolock

##### 1 Attachment

- geerky42

|dw:1433867445131:dw|

- geerky42

|dw:1433867464274:dw|
So \(\triangle ACD\) is equilateral triangle, right?

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## More answers

- geerky42

So can you tell me what \(m\angle BDA\) is?

- Zolock

I can not tell

- geerky42

You do agree that \(\triangle ACD\) is equilateral triangle, right?
So all angles in equilateral triangle are always ...?

- Zolock

yes it looks that way

- Zolock

they are alway equal

- geerky42

yeah. So since \(m\angle BDA\) is IN △ACD, what is angle?

- Zolock

BDA = 60 degress

- geerky42

|dw:1433867843613:dw|

- geerky42

Yeah, so there you go,

- Zolock

BDA is 60

- geerky42

|dw:1433867910800:dw|

- geerky42

Yeah. BDA is 60. Now what is DBA?

- Zolock

30

- geerky42

Correct. so you got answers for part A.

- geerky42

Now for part B, you have \(\sin(30^\text o)\)

- Zolock

so the sin for angle DBA is 30

- geerky42

Not quite. \(\sin(30^\text o)\neq 30\)

- geerky42

Do you have unit circle to check?

- Zolock

I did not get that

- geerky42

Ok, you know 30 60 90 triangle? |dw:1433868385520:dw|
So since \(\sin\theta\) is ratio of opposite side to hypotenuse side. In other words, \(\sin\theta = \dfrac{Opp}{Hyp}\), right?

- geerky42

We have Opp = a and Hyp = 2a. What is Opp/Hyp?

- Zolock

2

- geerky42

Not quite. Opp / Hyp = a / 2a
So it's 1/2

- Zolock

oh

- geerky42

\(\sin(30^\text o) = \dfrac{Opp}{Hyp} = \dfrac{a}{2a} = \dfrac{1}{2}\)
IS that clear?

- Zolock

yes

- Zolock

so sin(30 degrees) = 1/2

- geerky42

Yeah

- geerky42

Can you do part C?

- Zolock

nope I still need help with part C and part D I think I might me able to do part E on my own.

- geerky42

Ok. Whole circle is \(360^\text o\), right?|dw:1433869406908:dw|
Here, to find arc AB, we would need to find angle BCA.
|dw:1433869460152:dw|

- Zolock

120

- geerky42

Then we can find length of arc by using formula for circumference, kinda.
\(Arc = \dfrac{m\angle BCA}{360^\text o} ~~2\pi r\)

- geerky42

Here we have \(m\angle BCA = 120^\text o\)

- geerky42

So arc length BA is \(\dfrac{120^\text o}{360^\text o}~~~2\pi r = \dfrac{2\pi r}{3}\)
Make sense?

- geerky42

You were given that diameter is 18. So clearly radius is 9.

- Zolock

so the length of AB is 2πr/3

- geerky42

Yeah. now plug in r = 9

- Zolock

so AB is 2π^9/3

- Zolock

@geerky42 Can you finish helping me with part D and E when you get back online please I have to turn this in tomorrow.

- geerky42

π^9?
It's just \(\dfrac{2\pi 9}{3} = 2\pi3 = \boxed{6\pi}\)

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