help !!!

RauLiTo · 2006-11-19 00:13:46

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

mathsyperson · 2006-11-19 00:34:02

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°.

RauLiTo · 2006-11-19 00:42:04

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)

mathsyperson · 2006-11-19 01:51:09

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.

John E. Franklin · 2006-11-19 04:06:23

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??

mathsyperson · 2006-11-19 05:17:10

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.

RauLiTo · 2006-11-19 08:23:44

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

mathsyperson · 2006-11-19 09:07:39

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?

RauLiTo · 2006-11-19 23:18:10

thanks man i got it now thanks alot

Math Is Fun Forum

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

help !!!

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

Re: help !!!

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

Re: help !!!

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

Re: help !!!

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

Re: help !!!

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

Re: help !!!

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

Re: help !!!

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

Re: help !!!

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

Re: help !!!

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