# Math Is Fun Forum

Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫  π  -¹ ² ³ °

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## #1 2006-11-19 00:13:46

RauLiTo
Member
From: Bahrain
Registered: 2006-01-11
Posts: 142

### help !!!

what is the value of this :
(A) + (B) + (C) + (D) + (E)

ImPo\$\$!BLe = NoTH!nG
Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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## #2 2006-11-19 00:34:02

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: help !!!

So, we want to know the total of the indicated angles on that pentagram?

Well, let's look at A. A is part of the top triangle, so it will be 180° minus the other two angles.
The other two angles can combine with one of the angles in the inner pentagon to make a straight line.

So they are both 180° minus an angle on the inner pentagon. This means that angle A is 180° - ((180° - p) + (180° - q)), which is equal to p + q - 180° (where p and q are the appropriate angles in the pentagon).

Similar reasoning can be applied to B, C, D and E, and so if we label the inner angle of the pentagon as p, q, r, s and t, then the total of A, B, C, D and E is found by 2(p+q+r+s+t) - 900°.

We know that the angles in any pentagon will always total 540°, which means that the wanted angles add up to 2*540° - 900° = 180°.

Why did the vector cross the road?
It wanted to be normal.

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## #3 2006-11-19 00:42:04

RauLiTo
Member
From: Bahrain
Registered: 2006-01-11
Posts: 142

### Re: help !!!

sorry man but i i didnt get it all ! maybe because of my **** english ! can you explain more please and if its possible with drawing ?
thanks alot i am really embarrassed

Last edited by RauLiTo (2006-11-19 00:42:32)

ImPo\$\$!BLe = NoTH!nG
Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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## #4 2006-11-19 01:51:09

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: help !!!

I've added some more angles to the diagram to show what I'm talking about.

So, there's a triangle above the central pentagon made up of angles A, x and y. Because these make a triangle, we know that A + x + y = 180°.

However, angle x shares a straight line with angle p, and so x + p must also equal 180°. Rearranging gets us that x = 180° - p. Similar reasoning tells us that y = 180° - q.

So putting these values into the triangle's equation tells us that A + (180° - p) + (180° - q) = 180°. Rearranging for A gives that A = p + q - 180°.

Now we repeat this process for angles B, C, D and E.
The same reasoning will tell us that B = q + r - 180°, C = r + s - 180°, and so on.

Adding all of these tells us that A+B+C+D+E = 2(p+q+r+s+t) - 900°.

p+q+r+s+t is the total of all the angles in the pentagon, and we know that the total of angles in any pentagon is always 540°. Substituting this value in tells us that A+B+C+D+E= 2*540° - 900°, which is equal to 180°.

If you still don't follow, tell me which parts of my explanation you don't follow and I'll try to explain them better.

Why did the vector cross the road?
It wanted to be normal.

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## #5 2006-11-19 04:06:23

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

### Re: help !!!

That is simply amazing, mathyperson, Can you tell me how you knew the following:
and we know that the total of angles in any pentagon is always 540°.
How did you know that??

igloo myrtilles fourmis

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## #6 2006-11-19 05:17:10

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: help !!!

Thankyou very much!

I worked the pentagon fact out from the more general fact that the interior angles in a polygon with n sides will always total (n-2)*180°.

So, angles in a triangle add to (3-1)*180° = 180°, angles in a quadrilateral add to 360°, angles in a pentagon add to 540°, etc.

Why did the vector cross the road?
It wanted to be normal.

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## #7 2006-11-19 08:23:44

RauLiTo
Member
From: Bahrain
Registered: 2006-01-11
Posts: 142

### Re: help !!!

its really clear but i cant see the picture you put so i cant follow your steps

ImPo\$\$!BLe = NoTH!nG
Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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## #8 2006-11-19 09:07:39

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: help !!!

That's odd. I can see it fine, and presumably John could. Maybe imageshack was down for a while. Can you see it now (or when you next view this topic)?

And when you say that it's really clear, does that mean that you can follow it without a picture anyway?

Why did the vector cross the road?
It wanted to be normal.

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## #9 2006-11-19 23:18:10

RauLiTo
Member
From: Bahrain
Registered: 2006-01-11
Posts: 142

### Re: help !!!

thanks man i got it now thanks alot

ImPo\$\$!BLe = NoTH!nG
Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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