Liv1234
  • Liv1234
SOMEONE PLEASE HELP ME WITH A FEW QUESTIONS
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Liv1234
  • Liv1234
kc_kennylau
  • kc_kennylau
The tangent-secant theorem: \(CD^2 = CB \times CA\)
Liv1234
  • Liv1234
@kc_kennylau Can you help me through the problem?

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Liv1234
  • Liv1234
@kc_kennylau
Liv1234
  • Liv1234
Or anyone that can help please.
Godlovesme
  • Godlovesme
i'll help! :)
Godlovesme
  • Godlovesme
let's use the formula @kc_kennylau provided :3 \[CD^2 = BC+ AC\]
Godlovesme
  • Godlovesme
@Liv1234 u there?
Liv1234
  • Liv1234
Sorry, yes I'm here.
kc_kennylau
  • kc_kennylau
@Godlovesme that was times
Liv1234
  • Liv1234
Thank you for helping by the way.
Godlovesme
  • Godlovesme
oh sorry :/
kc_kennylau
  • kc_kennylau
it's alright
Liv1234
  • Liv1234
So, how would I use the formula to find out the answer?
Liv1234
  • Liv1234
@Godlovesme @kc_kennylau
kc_kennylau
  • kc_kennylau
Well, you have the length of \(CD\) and \(CB\)
Liv1234
  • Liv1234
So, 30*17?
kc_kennylau
  • kc_kennylau
Well, the formula is \(CD^2 = CB\times CA\)
Liv1234
  • Liv1234
Ohh, so 30^2=17*CA?
kc_kennylau
  • kc_kennylau
Yes :DDDD
Liv1234
  • Liv1234
A!
Liv1234
  • Liv1234
Can you help me with another question(:
kc_kennylau
  • kc_kennylau
No it's not A
kc_kennylau
  • kc_kennylau
A is actually the length of \(CA\)
kc_kennylau
  • kc_kennylau
You're asked to find the length of \(BA\)
Liv1234
  • Liv1234
Yes, it is because 30^2=900 and 17*52.9=900, right?
Liv1234
  • Liv1234
Ohh I see what you're saying sorry my bad.
kc_kennylau
  • kc_kennylau
it's alright xd
Liv1234
  • Liv1234
So, where did I mess up though?
kc_kennylau
  • kc_kennylau
Well, you only found the length of \(CA\).
kc_kennylau
  • kc_kennylau
How do you find the length of \(BA\) from that?
Liv1234
  • Liv1234
Divide by half?
kc_kennylau
  • kc_kennylau
Well, we can see that \(CA - CB = BA\)
Liv1234
  • Liv1234
Ohh, so 30-17?
kc_kennylau
  • kc_kennylau
Um.... no
kc_kennylau
  • kc_kennylau
Remember that \(CA = 52.9\)
Liv1234
  • Liv1234
Ohhh okay oopsies. cx
Liv1234
  • Liv1234
It's 35.9! Because 52.9-17=35.9
kc_kennylau
  • kc_kennylau
:D
kc_kennylau
  • kc_kennylau
yes :D
Liv1234
  • Liv1234
Can you help me again??
kc_kennylau
  • kc_kennylau
of course
Liv1234
  • Liv1234
kc_kennylau
  • kc_kennylau
Well, can you tell me \(m\angle OAB\)?
Liv1234
  • Liv1234
No, I don't understand it. :/ (and by the way I need help with like 3 other questions after this one if that's okay)
kc_kennylau
  • kc_kennylau
I mean the angle \(OAB\)
Liv1234
  • Liv1234
I don't know how to find it out that's what I'm confused about.
kc_kennylau
  • kc_kennylau
Well, since \(AB\) is tangent, the angle is actually \(90^\circ\)
Liv1234
  • Liv1234
Okay, so what do we do from there?
kc_kennylau
  • kc_kennylau
So, we can actually apply the Pythagoras' Theorem
Liv1234
  • Liv1234
Ohh okay. So, what was the formula for that?
kc_kennylau
  • kc_kennylau
\(OA^2+AB^2=OB^2\)
Liv1234
  • Liv1234
So, 24^2+AB^2=OB^2?
Liv1234
  • Liv1234
Did I put the numbers in wrong?
kc_kennylau
  • kc_kennylau
Nope
kc_kennylau
  • kc_kennylau
Can we find \(OC\)?
Liv1234
  • Liv1234
I think we can, but I'm not sure how.
kc_kennylau
  • kc_kennylau
Well, it is another radius
kc_kennylau
  • kc_kennylau
All radius have the same length
Liv1234
  • Liv1234
So, what would it be then?
Liv1234
  • Liv1234
I mean the radius.
Liv1234
  • Liv1234
Hello?
kc_kennylau
  • kc_kennylau
Well
kc_kennylau
  • kc_kennylau
\(OA\) is another radius
Liv1234
  • Liv1234
Ohh, so 90?
kc_kennylau
  • kc_kennylau
Em... what are we talking about?
Liv1234
  • Liv1234
The radius?
kc_kennylau
  • kc_kennylau
Well, the radius is 24
Liv1234
  • Liv1234
Oh okay, so OC would be 24?
kc_kennylau
  • kc_kennylau
Yes :D
kc_kennylau
  • kc_kennylau
So how long is \(OB\)?
Liv1234
  • Liv1234
24!
kc_kennylau
  • kc_kennylau
em.. i'm talking about \(OB\)
Liv1234
  • Liv1234
Ohh, oops.
Liv1234
  • Liv1234
I'm not sure what it would be, is there a way to find out?
kc_kennylau
  • kc_kennylau
Well, OB is just OC + CB
Liv1234
  • Liv1234
Would it be 48?
kc_kennylau
  • kc_kennylau
Nope. CB is actually 27
Liv1234
  • Liv1234
27+24=51
Liv1234
  • Liv1234
So, would the answer to the question be 51?
kc_kennylau
  • kc_kennylau
Nope, we need to find \(AB\)
Liv1234
  • Liv1234
And we'd do that by?
kc_kennylau
  • kc_kennylau
the pythagoras' theorem! :D
Liv1234
  • Liv1234
Ohh. And what was the formula again aha?
kc_kennylau
  • kc_kennylau
\(OA^2+AB^2=OB^2\)
Liv1234
  • Liv1234
So, 27^2+AB^2=51^2. 51^2=2601. Right?
Liv1234
  • Liv1234
I might have messed up with that a little.
kc_kennylau
  • kc_kennylau
it should be 24^2 instead of 27^2
Liv1234
  • Liv1234
Ohh okay. 24^2+AB^2=51^2.
kc_kennylau
  • kc_kennylau
:)
Liv1234
  • Liv1234
It would be 45. Which is A. Can you help me again?(:
Liv1234
  • Liv1234
kc_kennylau
  • kc_kennylau
Well, it's difficult to discuss without names
Liv1234
  • Liv1234
What do you mean?
kc_kennylau
  • kc_kennylau
Without names of vertices
kc_kennylau
  • kc_kennylau
I can't refer any line
Liv1234
  • Liv1234
Want me to post another question instead?
kc_kennylau
  • kc_kennylau
no
kc_kennylau
  • kc_kennylau
|dw:1434224619983:dw|
Liv1234
  • Liv1234
Oh okay, that's gonna make it a lot easier.
kc_kennylau
  • kc_kennylau
So, O is the centre of the circle
kc_kennylau
  • kc_kennylau
And OM makes a right angle with AB
kc_kennylau
  • kc_kennylau
So, AM=MB
kc_kennylau
  • kc_kennylau
(Just remember this result :p)
Liv1234
  • Liv1234
Okay. I will.
Liv1234
  • Liv1234
So, what do we start out with?
Liv1234
  • Liv1234
I'm confused.
kc_kennylau
  • kc_kennylau
So AM=MB
Liv1234
  • Liv1234
So, 21=5?
Liv1234
  • Liv1234
I mean 5=21?
kc_kennylau
  • kc_kennylau
Em...
kc_kennylau
  • kc_kennylau
I mean that M divides AB in half
Liv1234
  • Liv1234
So, half of 21?
kc_kennylau
  • kc_kennylau
yes, that would be MB
Liv1234
  • Liv1234
And half of 21 is 10.5
kc_kennylau
  • kc_kennylau
|dw:1434225065530:dw|
kc_kennylau
  • kc_kennylau
Yes
Liv1234
  • Liv1234
So, AM=10.5?
Liv1234
  • Liv1234
Right?
kc_kennylau
  • kc_kennylau
yes
Liv1234
  • Liv1234
Then what would we do?
kc_kennylau
  • kc_kennylau
Apply the Pythagoras' Theorem
Liv1234
  • Liv1234
Can you write it out for me again please?
kc_kennylau
  • kc_kennylau
\(OM^2+MB^2=OB^2\)
Liv1234
  • Liv1234
OM^2+10.5^2=OB^2?
kc_kennylau
  • kc_kennylau
You know OM
Liv1234
  • Liv1234
I do?
kc_kennylau
  • kc_kennylau
It's 5
Liv1234
  • Liv1234
5^2+10.5^2=OB^2?
kc_kennylau
  • kc_kennylau
yes
Liv1234
  • Liv1234
I got 135.25^2
kc_kennylau
  • kc_kennylau
yes
Liv1234
  • Liv1234
Which is 18,300.
kc_kennylau
  • kc_kennylau
that is OB^2
kc_kennylau
  • kc_kennylau
No, OB^2 is 135.25
Liv1234
  • Liv1234
I know, but when you do 135.25^2 you get 18300(rounded)
kc_kennylau
  • kc_kennylau
No, OB^2 is 135.25, so you root it, not square it
Liv1234
  • Liv1234
Okay.
Liv1234
  • Liv1234
Would it be A? 11.6?
kc_kennylau
  • kc_kennylau
yes
Liv1234
  • Liv1234
kc_kennylau
  • kc_kennylau
This is a theorem
Liv1234
  • Liv1234
What does that mean?
kc_kennylau
  • kc_kennylau
https://www.google.com/search?q=angle+of+centre+twice+angle+of+circumference&es_sm=93&tbm=isch&tbo=u&source=univ&sa=X&ved=0CCQQsARqFQoTCIe18qu7jcYCFah0pgodGiAA0w&biw=1093&bih=524&gws_rd=cr&ei=gIx8VYGANqbFmAWKpoPYBA
Liv1234
  • Liv1234
Can you help me with it?
kc_kennylau
  • kc_kennylau
Just times two
Liv1234
  • Liv1234
48?
kc_kennylau
  • kc_kennylau
The angle at center \(m\angle O\) is twice the angle at circumference \(m\angle R\) yes
Liv1234
  • Liv1234
Liv1234
  • Liv1234
What about this one?
Liv1234
  • Liv1234
And this is the last one.
kc_kennylau
  • kc_kennylau
Secant secant theorem: \(CA\times CB = CE \times CD\)
Liv1234
  • Liv1234
42*18=4*CD?
kc_kennylau
  • kc_kennylau
CA isn't 42
kc_kennylau
  • kc_kennylau
4 is CD, not CE
Liv1234
  • Liv1234
Okay, so 18*CB=CE*4?
Liv1234
  • Liv1234
CA*18=CE*4?
kc_kennylau
  • kc_kennylau
Yeah, how to find CA? It's the sum of two lines again
Liv1234
  • Liv1234
CA=64?
Liv1234
  • Liv1234
64*18=CE*4?
kc_kennylau
  • kc_kennylau
not 64, it's 60
Liv1234
  • Liv1234
60*18=CE*4?
kc_kennylau
  • kc_kennylau
yes
Liv1234
  • Liv1234
So, 270? Which is D?
Liv1234
  • Liv1234
Right?
kc_kennylau
  • kc_kennylau
no
kc_kennylau
  • kc_kennylau
You're asked to find DE
kc_kennylau
  • kc_kennylau
You need to deduct 4
Liv1234
  • Liv1234
1,066?
kc_kennylau
  • kc_kennylau
266
kc_kennylau
  • kc_kennylau
270-4=266
Liv1234
  • Liv1234
Ohh, that makes sense, so C?
Liv1234
  • Liv1234
?
kc_kennylau
  • kc_kennylau
yes
Liv1234
  • Liv1234
YAY thank you so much for helping(:
kc_kennylau
  • kc_kennylau
no problem

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