# Maths!!

Discussion in 'Community Discussion' started by cambookpro, Feb 7, 2012.

1. ### cambookpro macrumors 603

Joined:
Feb 3, 2010
Location:
United Kingdom
#1
Anyone else like maths?!

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?

File size:
60 KB
Views:
293
2. ### jav6454 macrumors P6

Joined:
Nov 14, 2007
Location:
1 Geostationary Tower Plaza
#2
All triangles add up to 180*

3. ### cambookpro thread starter macrumors 603

Joined:
Feb 3, 2010
Location:
United Kingdom
#3
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?

4. ### jav6454 macrumors P6

Joined:
Nov 14, 2007
Location:
1 Geostationary Tower Plaza
#4
Ok, let's start from the start....

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*

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

5. ### cambookpro thread starter macrumors 603

Joined:
Feb 3, 2010
Location:
United Kingdom
#5
Wow I'm such a moron, it's obvious now. Thanks for the help.
(That sounded sarcastic, honestly it's not!)

6. ### CalBoy macrumors 604

Joined:
May 21, 2007
#6
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.

7. ### jav6454 macrumors P6

Joined:
Nov 14, 2007
Location:
1 Geostationary Tower Plaza
#7

1. True

2. Doubt they are isosceles. But there are easier ways than trying to figure out if they are.

8. ### CalBoy macrumors 604

Joined:
May 21, 2007
#8
I must really obtuse right now, but how do you find Angle B using just OCD and the tangent line?

9. ### ender land macrumors 6502a

Joined:
Oct 26, 2010
#9
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.

10. ### CalBoy macrumors 604

Joined:
May 21, 2007
#10
I thought that OAD=OCD, but I couldn't remember why that was so (except if there were two isosceles triangles).

Joined:
Jan 25, 2009
12. ### jav6454 macrumors P6

Joined:
Nov 14, 2007
Location:
1 Geostationary Tower Plaza
#12
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.

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.

13. ### CalBoy macrumors 604

Joined:
May 21, 2007
#13
Of course you realize that if OCD=OAD, the two triangles are in fact isosceles.

You should post your proof(s) after the OP feels he has had the chance to work it out himself.

14. ### jav6454 macrumors P6

Joined:
Nov 14, 2007
Location:
1 Geostationary Tower Plaza
#14
I plan on it. But I'll wait till ~10PM Central.

15. Feb 7, 2012
Last edited: Feb 7, 2012

### samiwas macrumors 68000

Joined:
Aug 26, 2006
Location:
Atlanta, GA
#15
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.

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.

It's an hour later than that...come on!!!

16. ### jav6454 macrumors P6

Joined:
Nov 14, 2007
Location:
1 Geostationary Tower Plaza
#16

Part A

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*

17. Feb 7, 2012
Last edited: Feb 8, 2012

### iStudentUK macrumors 65816

Joined:
Mar 8, 2009
Location:
London
#17
How do we know triangle ADC is isosceles?

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.

18. ### eawmp1 macrumors 601

Joined:
Feb 19, 2008
Location:
FL
#18
Without spoon-feeding, this PDF on circle theorem proofs will help.

19. ### cambookpro thread starter macrumors 603

Joined:
Feb 3, 2010
Location:
United Kingdom
#19
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

20. ### jav6454 macrumors P6

Joined:
Nov 14, 2007
Location:
1 Geostationary Tower Plaza
#20
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.

Anyways, hope I helped OP

21. ### lewis82 macrumors 68000

Joined:
Aug 26, 2009
Location:
Totalitarian Republic of Northlandia
#21
Wow - this is so far away in my head! I could have solved OP's problem, but not some more complicated geometry stuff.

I can help people with triple integrals, though

22. Feb 8, 2012
Last edited: Feb 8, 2012

### samiwas macrumors 68000

Joined:
Aug 26, 2006
Location:
Atlanta, GA
#22
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.

23. ### camardelle macrumors 6502

Joined:
Aug 17, 2011
Location:
Texas
#23
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.

24. ### ender land macrumors 6502a

Joined:
Oct 26, 2010
#24

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..

25. ### hakuna-matata macrumors 6502

Joined:
Sep 25, 2011
#25
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