http://gyazo.com/f6cd34893dc845eca0d0a61500763ce5
whats up
next time post the question first
The question IS posted there @triciaal
is this goemetry
Yep
sorry cant help you with this one i have not learned it yet
Ah man, this is due tonight. Anyone else that can maybe help :p ?
hint : \[\large \text{inscribed angle} = \dfrac{\text{arc measure}}{2}\]
find the angles ADE and BAD first
all paraphraased the angle at center is directly proportional to the length of the arc
So @ganeshie8 it would be either A or B right?
Im not sure which question are we working on
I see you have attached two links and they have different questions
could you clarify
if the center = O the angle ACE = 1/2 AOE
oh the second like is just the measure (ignore the text on top of it). The original screenshot is the one I'm working on. With the A, B, C and D options. @ganeshie8
@triciaal Their is no option for ACE = 1/2(AOE)
still i dont get it, could you put everything in one picture and post it ?
Yea sorry one second
Where it says "in the diagram above". This is the Diagram it is talking about: http://gyazo.com/5c19372ed3b430ce2387dab9775da392
@CoolGuy13 I was just working through "aloud" I don't have the actual answer yet and I can't see your options. I am not very fast with these either. just trying to help. as expressed by @ganeshie8 as well given the arc length find the angles. you have 45 and 110
@triciaal Thanks for helping!
@ganeshie8 Do you see the new problem?
Alright, so you're working on Part II, question a ?
Yes
Okay good, there is a formula for finding the angle ACE outside a point : \[\large m\angle ACE = \dfrac{\text{far arc - near arc}}{2}\]
look at the given diagram from angle ACE
far arc = ? near arc = ?
far arc = AE near arc = BD
perfect ! so plug them in
\[\large m\angle ACE = \dfrac{\text{mAE - mBD}}{2} \]
So answer would be B ?
Yep!
32.5 for part B?
AE = 110 BD = 45 110 - 45 = 65 65 / 2 = 32.5 m<ACE = 32.5
And for Part C, how would i use this to find M<BAD ? @ganeshie8
m<ACE = 32.5 is \(\large \color{Red}{\checkmark}\)
use inscribed angle formula for part C
\[\large \text{inscribed angle} = \dfrac{\text{arc measure}}{2}\]
What would arc measure be?
BAD eats the arc 45 right ?
Yea
45 / 2 ?
Yep!
thank you so much @ganeshie8. I have 2 (a lot shorter) more problems and this is due in about an hour. Think you can help me out with them?
sure post them here
Would part 1 be 360 - 90 ?
@ganeshie8 ?
Part I : \[\large \dfrac{90}{360}\]
which equals 1/4 right ?
basically the length of arc AB would be 1/4 of the circumference of the full circle
knw how to work part II ?
emmm
Can you refresh my memory?
diameter = 32 radius = ?
circumference = \(\large 2~\pi ~r = ?\)
Wait isn't the equation right on top of the figure?
yes
you have already calculated mAB/360 = 90/360 = 1/4
find the circumference and multiply that by 1/4
2 * 3.14 * 16 = 100.48 * 1/4 = 25.12
Is that right?
looks good!
Ok phew one last question left, i really appreciate all the help man (:. http://gyazo.com/5328a5976af86524e1292cc261061266
If that's ok with you
diameter = 20 radius = ?
radius = 10
|dw:1414131147632:dw|
Full circle = 360 if you cut it into 20 pieces like above, what would be the angle at center ?
|dw:1414131241869:dw|
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