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