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Maths!!

cambookpro

cambookpro

macrumors 604
Original poster
Feb 3, 2010
7,235
3,387
United Kingdom
Anyone else like maths?! :p

Anyway, I was wondering if anyone could point me in the direction (Read: I don't just want the answer on a plate) of solving this (attached).

Obviously RCB is 90-34=56, but from there I have no idea where to go. I was thinking that in a cyclic quadrilateral opposite vertices add up to 180 degrees, but then of course OCDA isn't cyclic.

Once you get ABC then OAD should be easy, just 360-(2ABC)-137-72.

Any ideas?
 

Attachments

  • Screen Shot 2012-02-07 at 21.42.39.png
    Screen Shot 2012-02-07 at 21.42.39.png
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cambookpro said:
I did know that, however I don't follow where you apply this?
Are you saying in triADC that angles A and C are (180-137)/2?
Click to expand...

Ok, let's start from the start....

First Answer

Line RCT is a tangent line. to the circle. By definition a tanglent line is one that touches the circle at one point and is at a 90* angle from that point with respect to the radii line. In other words, RCT forms a 90* with line segment OC.

Given:

Angle OCB is 34*, which means angle RCB is 90* minus the angle from OCB. That is 90-34 which yields 56*

Second Answer

With what I told you in the first, the other two are extremely easy....
 
jav6454 said:
Ok, let's start from the start....

First Answer

Line RCT is a tangent line. to the circle. By definition a tanglent line is one that touches the circle at one point and is at a 90* angle from that point with respect to the radii line. In other words, RCT forms a 90* with line segment OC.

Given:

Angle OCB is 34*, which means angle RCB is 90* minus the angle from OCB. That is 90-34 which yields 56*

Second Answer

With what I told you in the first, the other two are extremely easy....
Click to expand...

Wow I'm such a moron, it's obvious now. Thanks for the help. :)
(That sounded sarcastic, honestly it's not!)
 
Angle OCD should be 62 (90-28) based on the fact that the line segment is a tangent.

Unless I'm mistaken, the two smaller triangles are isosceles, which should help you out a great deal.
 
CalBoy said:
Angle OCD should be 62 (90-28) based on the fact that the line segment is a tangent.

Unless I'm mistaken, the two smaller triangles are isosceles, which should help you out a great deal.
Click to expand...


1. True

2. Doubt they are isosceles. But there are easier ways than trying to figure out if they are.
 
jav6454 said:
2. Doubt they are isosceles. But there are easier ways than trying to figure out if they are.
Click to expand...

I must really obtuse right now, but how do you find Angle B using just OCD and the tangent line?
 
CalBoy said:
I must really obtuse right now, but how do you find Angle B using just OCD and the tangent line?
Click to expand...

Acutely speaking, you know angle OCT = 90.

This lets you find OCD = 90-28, which is the same as OAD.

OAD + OCD + 128 + B = 360.


Also, you know triangle OAC is isosceles because two of its legs are radii. You do NOT know whether OCD is as well.
 
ender land said:
Acutely speaking, you know angle OCT = 90.

This lets you find OCD = 90-28, which is the same as OAD.

OAD + OCD + 128 + B = 360.


Also, you know triangle OAC is isosceles because two of its legs are radii. You do NOT know whether OCD is as well.
Click to expand...

I thought that OAD=OCD, but I couldn't remember why that was so (except if there were two isosceles triangles).
 
CalBoy said:
I thought that OAD=OCD, but I couldn't remember why that was so (except if there were two isosceles triangles).
Click to expand...

Can't make those assumptions. All we are given are three angles, nothing more. We are not told what segments are congruent or equal. Hence, we can't base this entire problem on using isosceles triangles and so forth.

LostSoul80 said:
Hint: COA = 2*CBA.
Click to expand...

No, this is not a hint, this is a death sentence. You can't make this assumption based on the fact that you don't know the length of those segments. All we can go on are the angles.




People, this problem is *very* easy. It took me 5 minutes (TOPS) to go through the whole thing.
 
jav6454 said:
Can't make those assumptions. All we are given are three angles, nothing more. We are not told what segments are congruent or equal. Hence, we can't base this entire problem on using isosceles triangles and so forth.
Click to expand...

Of course you realize that if OCD=OAD, the two triangles are in fact isosceles.

People, this problem is *very* easy. It took me 5 minutes (TOPS) to go through the whole thing.
Click to expand...

You should post your proof(s) after the OP feels he has had the chance to work it out himself.
 
I'm usually pretty damn good with trig and geometry, but this one has me a bit stumped. I'll be interested in jav's response.

CalBoy said:
Of course you realize that if OCD=OAD, the two triangles are in fact isosceles.
Click to expand...

We do not know that OCD=OAD. We know that OAC is an isosceles triangle, but that's about it. We know that OCD is 62˚, but I'm not sure how you divide that up for the two angles within that.

I mean, I can give you the answers, but I'm using CAD to figure it out!

EDIT: Interesting...so when I dumped this into CAD, I found that the way it's drawn is not what reality is. The line OA has a downward slope in reality, not an upward slope. So it's definitely tricking your mind just to look at it and assume anything.


jav6454 said:
I plan on it. But I'll wait till ~10PM Central.
Click to expand...

It's an hour later than that...come on!!!
 
Last edited:
samiwas said:
It's an hour later than that...come on!!!
Click to expand...

Ooops, forgot about this thread... anyways...

Answer
Part A

Already given above.

Part B

We know triangle ADC has one angle at 137*, that means there is only 43* left for the other two angles. This means, each angle is at 21.5* each.

Now, we know that angle OCD is 90-28 which is 62*. However, we are looking for ABC. To know triangle ABC, I must know one of the angles of it. In this case its the angle at C using triangle ABC as reference. I already know that 62* are left from the previous. So I know subtract the difference from triangle ACD and get 62-21.5 = 40.5*. Now add this 40.5 to the 34* we are given we obtain a 74.5* angle. This is the same angle at the other side, meaning we've got two angles found for triangle ABC. This means we have accounted for 149* in this triangle. Now I need the angle of B taking triangle ABC into consideration. Easy... 180-149 that means the angle of B using ABC is 31*.

Part C

Using from the previous parts... the answer for OAD is the same answer for OCD which boils down to 62*
 
jav6454 said:
We know triangle ADC has one angle at 137*, that means there is only 43* left for the other two angles. This means, each angle is at 21.5* each.
Click to expand...

How do we know triangle ADC is isosceles? :confused:

Doesn't segment theorem suggest CAD=TCD=28*

This means DCA=180-28-137=15*

OCA=90-28-15=47*

Again segment theorem says CAB=RCB=56*

So ABC=180-56-34-47=43*

Am I missing something?

EDIT- or even quicker once you have DCT and ACD just add them together (28+15=43) and that gives you ABC, again y segment theorem.
 
Last edited:
Thanks for your help everyone, especially jav.
I'm usually OK at maths, quadratic simultaneous equations, sine and cosine rule, completing the square, plotting inequalities etc, however this question had me stumped.

I'm only in year 9 (age 14) so will remember this for my GCSEs and A levels :p
 
iStudentUK said:
How do we know triangle ADC is isosceles? :confused:

Doesn't segment theorem suggest CAD=TCD=28*

This means DCA=180-28-137=15*

OCA=90-28-15=47*

Again segment theorem says CAB=RCB=56*

So ABC=180-56-34-47=43*

Am I missing something?

EDIT- or even quicker once you have DCT and ACD just add them together (28+15=43) and that gives you ABC, again y segment theorem.
Click to expand...

I assumed it was. I did however warn it may very well not be isosceles. However, that is my logic thinking within 5 minutes. I did don't go deep into the problem. I know there must be a way, but I am here to give a help towards the answer, not the answer. :p

Anyways, hope I helped OP :)
 
iStudentUK said:
How do we know triangle ADC is isosceles? :confused:

Doesn't segment theorem suggest CAD=TCD=28*

This means DCA=180-28-137=15*

OCA=90-28-15=47*

Again segment theorem says CAB=RCB=56*

So ABC=180-56-34-47=43*

Am I missing something?

EDIT- or even quicker once you have DCT and ACD just add them together (28+15=43) and that gives you ABC, again y segment theorem.
Click to expand...

All of these numbers are correct. I don't know anything about segment theorem, but I will now be looking that up, because it sounds interesting. Seems like any two segments coming off a tangent line will produce an identical angle back to the originating point. That's kind of fascinating. I think that is the key to figuring this whole thing out.
 
Last edited:
Where were you guys when I was in high school (probably not born yet I'm thinking). I would have been sitting next to you during math. :D
 
jav6454 said:
I assumed it was. I did however warn it may very well not be isosceles. However, that is my logic thinking within 5 minutes. I did don't go deep into the problem. I know there must be a way, but I am here to give a help towards the answer, not the answer. :p

Anyways, hope I helped OP :)
Click to expand...


You know, if you are going to act smug about having only spent 5 minutes on a math problem, you probably should do the problem correctly..
 
IS IT SOLVED YET??? sry didn't read the above ans..they were so many..just tel me if it is solved yet or not?..i will give it a try then
 
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