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How many pairs of vertical angles are formed by five distinct lines that have a common point of intersection?

 

I got n^2=25

 

But it was wrong...

 

How do I get it right?

 

Can somebody please help me out?

 Jul 15, 2019
 #1
avatar+103122 
+3

These angles will have a corresponding pair  

AKE , IKA, GKI, DKG , FKD

EKI, AKG, IKD, GKF, DKB

EKG, AKD IKF, GKB , DKJ

EKD , AKF, IKB, GKJ, DKH

EKF, AKB, IKJ, GKH,DKC

 

I get  n^2  = 25, as well, tom......

 

 

cool cool cool

 Jul 16, 2019
 #2
avatar+1342 
+6

Yeah is there another answer you got? before rom corrected you last time?

 

I saw your answer on mellies... I thought it made sense, but now i need to get it right so I can look at the solution.

 

And basically figure it out...

tommarvoloriddle  Jul 16, 2019
edited by tommarvoloriddle  Jul 16, 2019
 #3
avatar+103122 
0

My first answer was incorrect, tom......maybe someone else has another approach/answer

 

 

cool cool cool

CPhill  Jul 16, 2019
 #4
avatar+1342 
0

yeah but what was it? my first one was 10.

tommarvoloriddle  Jul 16, 2019
 #5
avatar+103122 
0

Sorry....I don't remember.....!!!

 

cool cool cool

CPhill  Jul 16, 2019
 #6
avatar+1342 
0

:( I wish i got it right though- I treid counting and i got alot more that 30 but i stopped- it was 2 messy...

tommarvoloriddle  Jul 16, 2019
 #7
avatar+1342 
0

What about angles such as CKJ? you left that out... We are nvr going to finish counting...

😭😭😭😭😭😭😭😭😭😭

tommarvoloriddle  Jul 16, 2019
 #8
avatar+103122 
0

Angle CKJ is paired wth angle IKD

 

 

cool cool cool

CPhill  Jul 16, 2019
 #9
avatar+1342 
0

so isnt it 50?

tommarvoloriddle  Jul 16, 2019
 #10
avatar+103122 
0

No...we have  25 pairs  (by my count)  ....for example CKJ and IKD  is the same pair as IKD and CKJ

 

 

 

cool cool cool 

CPhill  Jul 16, 2019
 #11
avatar+1342 
0

But I think its 50. Because you should count every angle, and as long as they have different names, I think it counts. Not sure...

tommarvoloriddle  Jul 16, 2019
 #12
avatar+771 
+5

ibelive there is 27 pairs sorry but thats what i got i may be wrong though because i did it off th example of cp's pic and tried my best

travisio  Jul 17, 2019
edited by travisio  Jul 17, 2019
 #13
avatar+1342 
0

Ok it's fine travis

tommarvoloriddle  Jul 18, 2019
 #14
avatar+1699 
+1

 

Solution: 

When two (2) distinct lines intersect, four (4) vertical angles are formed.

For five lines:

The first line forms 4 angles with each of the four other lines.  4(4) = 16

The second line forms 4 angles with each of the three other lines.  3(4) = 12

The third line forms 4 angles with each of the two other lines.  2(4) = 8

The fourth line forms 4 angles with the remaining line.  1(4)  = 4

 

There are 16 + 12 + 8 + 4 = 40 vertical angles.

 

 

GA

 Jul 18, 2019
edited by GingerAle  Jul 18, 2019
edited by GingerAle  Jul 18, 2019
 #19
avatar+1699 
+1

I thought it strange that all these angles should be vertical. Logically, half should be “horizontal” angles.

 

Below is an image excerpt from Methods for Euclidean Geometry –37th Ed.  (Copyright 2010)

 

 

The text above the theorem [seems to] indicate[s] all the angles are vertical.

 

The theorem itself indicates only two are vertical.  It doesn’t indentify the other angles, only that they are also congruent. 

 

Next time, I’ll divide by 2. 

 

GA

GingerAle  Jul 20, 2019
 #20
avatar+103689 
0

Well you will have tje same answer as us then anyway.    cheeky

Melody  Jul 20, 2019
 #15
avatar+103122 
+5

Corrected answer  [ I think ]

 

I was  including "straight angle"  pairs.....but.....this is not  correct.....

 

The Vertical Angles Theorem states that vertical angles, angles that are opposite each other and formed by two intersecting lines, are congruent. 

 

So...."straight angle" pairs should not included

 

So.....the correct answer - I believe -   is just 20 pairs of vertical angles ......

 

 

 

cool cool cool

 Jul 19, 2019
 #21
avatar+1342 
0

You think?

tommarvoloriddle  Jul 22, 2019
 #16
avatar+103689 
+4

I got

1+1=2 for 2 lines

2+2+2=6 for 3 lines

3+3+3+3=12 for 4 lines

4+4+4+4+4=20 for 5 lines

 

and by extension

(n-1)n    for n lines.

 

So Chris we are in agreement. 

 

Tom, can you see if 20 is accepted as the correct answer?

 Jul 19, 2019
 #17
avatar+103122 
+2

THX, Melody.....!!!

 

 

cool cool cool

CPhill  Jul 19, 2019
 #18
avatar+1342 
+7

ok the answer is 20... i was wrong... :(

tommarvoloriddle  Jul 19, 2019

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