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?
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......
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...
My first answer was incorrect, tom......maybe someone else has another approach/answer
:( I wish i got it right though- I treid counting and i got alot more that 30 but i stopped- it was 2 messy...
What about angles such as CKJ? you left that out... We are nvr going to finish counting...
No...we have 25 pairs (by my count) ....for example CKJ and IKD is the same pair as IKD and CKJ
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...
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.
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.
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 ......
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?