SOMEONE PLEASE HELP ME WITH A FEW QUESTIONS

- Liv1234

SOMEONE PLEASE HELP ME WITH A FEW QUESTIONS

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- Liv1234

##### 1 Attachment

- kc_kennylau

The tangent-secant theorem: \(CD^2 = CB \times CA\)

- Liv1234

@kc_kennylau Can you help me through the problem?

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## More answers

- Liv1234

@kc_kennylau

- Liv1234

Or anyone that can help please.

- Godlovesme

i'll help! :)

- Godlovesme

let's use the formula @kc_kennylau provided :3
\[CD^2 = BC+ AC\]

- Godlovesme

@Liv1234 u there?

- Liv1234

Sorry, yes I'm here.

- kc_kennylau

@Godlovesme that was times

- Liv1234

Thank you for helping by the way.

- Godlovesme

oh sorry :/

- kc_kennylau

it's alright

- Liv1234

So, how would I use the formula to find out the answer?

- Liv1234

@Godlovesme @kc_kennylau

- kc_kennylau

Well, you have the length of \(CD\) and \(CB\)

- Liv1234

So, 30*17?

- kc_kennylau

Well, the formula is \(CD^2 = CB\times CA\)

- Liv1234

Ohh, so 30^2=17*CA?

- kc_kennylau

Yes :DDDD

- Liv1234

A!

- Liv1234

Can you help me with another question(:

- kc_kennylau

No it's not A

- kc_kennylau

A is actually the length of \(CA\)

- kc_kennylau

You're asked to find the length of \(BA\)

- Liv1234

Yes, it is because 30^2=900 and 17*52.9=900, right?

- Liv1234

Ohh I see what you're saying sorry my bad.

- kc_kennylau

it's alright xd

- Liv1234

So, where did I mess up though?

- kc_kennylau

Well, you only found the length of \(CA\).

- kc_kennylau

How do you find the length of \(BA\) from that?

- Liv1234

Divide by half?

- kc_kennylau

Well, we can see that \(CA - CB = BA\)

- Liv1234

Ohh, so 30-17?

- kc_kennylau

Um.... no

- kc_kennylau

Remember that \(CA = 52.9\)

- Liv1234

Ohhh okay oopsies. cx

- Liv1234

It's 35.9! Because 52.9-17=35.9

- kc_kennylau

:D

- kc_kennylau

yes :D

- Liv1234

Can you help me again??

- kc_kennylau

of course

- Liv1234

##### 1 Attachment

- kc_kennylau

Well, can you tell me \(m\angle OAB\)?

- 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

I mean the angle \(OAB\)

- Liv1234

I don't know how to find it out that's what I'm confused about.

- kc_kennylau

Well, since \(AB\) is tangent, the angle is actually \(90^\circ\)

- Liv1234

Okay, so what do we do from there?

- kc_kennylau

So, we can actually apply the Pythagoras' Theorem

- Liv1234

Ohh okay. So, what was the formula for that?

- kc_kennylau

\(OA^2+AB^2=OB^2\)

- Liv1234

So, 24^2+AB^2=OB^2?

- Liv1234

Did I put the numbers in wrong?

- kc_kennylau

Nope

- kc_kennylau

Can we find \(OC\)?

- Liv1234

I think we can, but I'm not sure how.

- kc_kennylau

Well, it is another radius

- kc_kennylau

All radius have the same length

- Liv1234

So, what would it be then?

- Liv1234

I mean the radius.

- Liv1234

Hello?

- kc_kennylau

Well

- kc_kennylau

\(OA\) is another radius

- Liv1234

Ohh, so 90?

- kc_kennylau

Em... what are we talking about?

- Liv1234

The radius?

- kc_kennylau

Well, the radius is 24

- Liv1234

Oh okay, so OC would be 24?

- kc_kennylau

Yes :D

- kc_kennylau

So how long is \(OB\)?

- Liv1234

24!

- kc_kennylau

em.. i'm talking about \(OB\)

- Liv1234

Ohh, oops.

- Liv1234

I'm not sure what it would be, is there a way to find out?

- kc_kennylau

Well, OB is just OC + CB

- Liv1234

Would it be 48?

- kc_kennylau

Nope. CB is actually 27

- Liv1234

27+24=51

- Liv1234

So, would the answer to the question be 51?

- kc_kennylau

Nope, we need to find \(AB\)

- Liv1234

And we'd do that by?

- kc_kennylau

the pythagoras' theorem! :D

- Liv1234

Ohh. And what was the formula again aha?

- kc_kennylau

\(OA^2+AB^2=OB^2\)

- Liv1234

So, 27^2+AB^2=51^2. 51^2=2601. Right?

- Liv1234

I might have messed up with that a little.

- kc_kennylau

it should be 24^2 instead of 27^2

- Liv1234

Ohh okay. 24^2+AB^2=51^2.

- kc_kennylau

:)

- Liv1234

It would be 45. Which is A. Can you help me again?(:

- Liv1234

##### 1 Attachment

- kc_kennylau

Well, it's difficult to discuss without names

- Liv1234

What do you mean?

- kc_kennylau

Without names of vertices

- kc_kennylau

I can't refer any line

- Liv1234

Want me to post another question instead?

- kc_kennylau

no

- kc_kennylau

|dw:1434224619983:dw|

- Liv1234

Oh okay, that's gonna make it a lot easier.

- kc_kennylau

So, O is the centre of the circle

- kc_kennylau

And OM makes a right angle with AB

- kc_kennylau

So, AM=MB

- kc_kennylau

(Just remember this result :p)

- Liv1234

Okay. I will.

- Liv1234

So, what do we start out with?

- Liv1234

I'm confused.

- kc_kennylau

So AM=MB

- Liv1234

So, 21=5?

- Liv1234

I mean 5=21?

- kc_kennylau

Em...

- kc_kennylau

I mean that M divides AB in half

- Liv1234

So, half of 21?

- kc_kennylau

yes, that would be MB

- Liv1234

And half of 21 is 10.5

- kc_kennylau

|dw:1434225065530:dw|

- kc_kennylau

Yes

- Liv1234

So, AM=10.5?

- Liv1234

Right?

- kc_kennylau

yes

- Liv1234

Then what would we do?

- kc_kennylau

Apply the Pythagoras' Theorem

- Liv1234

Can you write it out for me again please?

- kc_kennylau

\(OM^2+MB^2=OB^2\)

- Liv1234

OM^2+10.5^2=OB^2?

- kc_kennylau

You know OM

- Liv1234

I do?

- kc_kennylau

It's 5

- Liv1234

5^2+10.5^2=OB^2?

- kc_kennylau

yes

- Liv1234

I got 135.25^2

- kc_kennylau

yes

- Liv1234

Which is 18,300.

- kc_kennylau

that is OB^2

- kc_kennylau

No, OB^2 is 135.25

- Liv1234

I know, but when you do 135.25^2 you get 18300(rounded)

- kc_kennylau

No, OB^2 is 135.25, so you root it, not square it

- Liv1234

Okay.

- Liv1234

Would it be A? 11.6?

- kc_kennylau

yes

- Liv1234

##### 1 Attachment

- kc_kennylau

This is a theorem

- Liv1234

What does that mean?

- 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

Can you help me with it?

- kc_kennylau

Just times two

- Liv1234

48?

- kc_kennylau

The angle at center \(m\angle O\) is twice the angle at circumference \(m\angle R\)
yes

- Liv1234

##### 1 Attachment

- Liv1234

What about this one?

- Liv1234

And this is the last one.

- kc_kennylau

Secant secant theorem:
\(CA\times CB = CE \times CD\)

- Liv1234

42*18=4*CD?

- kc_kennylau

CA isn't 42

- kc_kennylau

4 is CD, not CE

- Liv1234

Okay, so 18*CB=CE*4?

- Liv1234

CA*18=CE*4?

- kc_kennylau

Yeah, how to find CA? It's the sum of two lines again

- Liv1234

CA=64?

- Liv1234

64*18=CE*4?

- kc_kennylau

not 64, it's 60

- Liv1234

60*18=CE*4?

- kc_kennylau

yes

- Liv1234

So, 270? Which is D?

- Liv1234

Right?

- kc_kennylau

no

- kc_kennylau

You're asked to find DE

- kc_kennylau

You need to deduct 4

- Liv1234

1,066?

- kc_kennylau

266

- kc_kennylau

270-4=266

- Liv1234

Ohh, that makes sense, so C?

- Liv1234

?

- kc_kennylau

yes

- Liv1234

YAY thank you so much for helping(:

- kc_kennylau

no problem

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