Zolock
  • Zolock
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:
Mathematics
katieb
  • katieb
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Zolock
  • Zolock
1 Attachment
geerky42
  • geerky42
|dw:1433867445131:dw|
geerky42
  • geerky42
|dw:1433867464274:dw| So \(\triangle ACD\) is equilateral triangle, right?

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geerky42
  • geerky42
So can you tell me what \(m\angle BDA\) is?
Zolock
  • Zolock
I can not tell
geerky42
  • geerky42
You do agree that \(\triangle ACD\) is equilateral triangle, right? So all angles in equilateral triangle are always ...?
Zolock
  • Zolock
yes it looks that way
Zolock
  • Zolock
they are alway equal
geerky42
  • geerky42
yeah. So since \(m\angle BDA\) is IN △ACD, what is angle?
Zolock
  • Zolock
BDA = 60 degress
geerky42
  • geerky42
|dw:1433867843613:dw|
geerky42
  • geerky42
Yeah, so there you go,
Zolock
  • Zolock
BDA is 60
geerky42
  • geerky42
|dw:1433867910800:dw|
geerky42
  • geerky42
Yeah. BDA is 60. Now what is DBA?
Zolock
  • Zolock
30
geerky42
  • geerky42
Correct. so you got answers for part A.
geerky42
  • geerky42
Now for part B, you have \(\sin(30^\text o)\)
Zolock
  • Zolock
so the sin for angle DBA is 30
geerky42
  • geerky42
Not quite. \(\sin(30^\text o)\neq 30\)
geerky42
  • geerky42
Do you have unit circle to check?
Zolock
  • Zolock
I did not get that
geerky42
  • 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
  • geerky42
We have Opp = a and Hyp = 2a. What is Opp/Hyp?
Zolock
  • Zolock
2
geerky42
  • geerky42
Not quite. Opp / Hyp = a / 2a So it's 1/2
Zolock
  • Zolock
oh
geerky42
  • geerky42
\(\sin(30^\text o) = \dfrac{Opp}{Hyp} = \dfrac{a}{2a} = \dfrac{1}{2}\) IS that clear?
Zolock
  • Zolock
yes
Zolock
  • Zolock
so sin(30 degrees) = 1/2
geerky42
  • geerky42
Yeah
geerky42
  • geerky42
Can you do part C?
Zolock
  • 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
  • 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
  • Zolock
120
geerky42
  • 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
  • geerky42
Here we have \(m\angle BCA = 120^\text o\)
geerky42
  • geerky42
So arc length BA is \(\dfrac{120^\text o}{360^\text o}~~~2\pi r = \dfrac{2\pi r}{3}\) Make sense?
geerky42
  • geerky42
You were given that diameter is 18. So clearly radius is 9.
Zolock
  • Zolock
so the length of AB is 2πr/3
geerky42
  • geerky42
Yeah. now plug in r = 9
Zolock
  • Zolock
so AB is 2π^9/3
Zolock
  • 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
  • geerky42
π^9? It's just \(\dfrac{2\pi 9}{3} = 2\pi3 = \boxed{6\pi}\)

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