Formula for angle in Polygon: size of angle ACD in terms of n

Simonsky

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Question: ABCDE.......is a regular polygon with n sides. Write an expression for the size of angle ACD in terms of n.


So I drew this:Polygon formula.jpg

This means the angle at C can be worked out as the interior angle minus one of the base angles of the isosceles triangle formed by the line AC. So the formula for the interior angle is: (n-2)180/n and the base angle will be 0.5[180-(n-2)180/n]

So I put that together to get: ACD = (n-2)180/n - 0.5[180 -(n-2)180/n] am I on the write track here? And can anyone help me simplify this cumbersome expression?

Many thanks, as always for generous help.
 
Question: ABCDE.......is a regular polygon with n sides. Write an expression for the size of angle ACD in terms of n.


So I drew this:View attachment 9534

This means the angle at C can be worked out as the interior angle minus one of the base angles of the isosceles triangle formed by the line AC. So the formula for the interior angle is: (n-2)180/n and the base angle will be 0.5[180-(n-2)180/n]

So I put that together to get: ACD = (n-2)180/n - 0.5[180 -(n-2)180/n] am I on the write track here? And can anyone help me simplify this cumbersome expression?

Many thanks, as always for generous help.

In my opinion, this is one of the most useless formulas ever.

The sum of the EXTERNAL angles of polygon. is always 360º.
For a Regular n-gon each EXTERNAL angle measures 360º / n.
Thus, each INTERNAL angle measures [180º - (360º / n)] = 180º * (1 - (2/n)). Quite a bit cleaner.
 
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Question: ABCDE.......is a regular polygon with n sides. Write an expression for the size of angle ACD in terms of n.

So I drew this:

This means the angle at C can be worked out as the interior angle minus one of the base angles of the isosceles triangle formed by the line AC. So the formula for the interior angle is: (n-2)180/n and the base angle will be 0.5[180-(n-2)180/n]

So I put that together to get: ACD = (n-2)180/n - 0.5[180 -(n-2)180/n] am I on the write track here? And can anyone help me simplify this cumbersome expression?.

Yes, your work is correct so far. Now you can just expand each parenthesis in turn. It's long, but not difficult. If you want more help with simplifying it, show us your attempt and we can identify your mistakes, if any.

It can also be helpful to simplify as you go: simplify 0.5[180-(n-2)180/n] before using it.

I got my answer without going through that, using the simpler formula that tkhunny referred to: The external angle is a = 360/n. The base angles of the isosceles triangle are a/2; so the supplement of ACD is a + a/2 = 3a/2. Write an expression for the supplement of this; then replace a with 360/n and simplify.

Note how using temporary variables can keep things from getting too complicated. Doing it your way, you could define x = (n-2)180/n and write everything in terms of that, simplifying before you replace x with its expression.
 
This is the bit I find difficult:

So: [180(n-2)]/2 =(180n-360)/n then: 1/2[ 180- (180n -360)/n)]

So that leaves me (180n-360)/n -1/2[180-(180n-360)/n]

Not sure how to deal with second bracket but here goes: (90 +(90n-180)/n)

so that leaves: (180n-360)/n - 90 + (90n - 180/n)


Hmmm......think I'm losing the plot now....
 
This is the bit I find difficult:

So: [180(n-2)]/2 =(180n-360)/n then: 1/2[ 180- (180n -360)/n)]

So that leaves me (180n-360)/n -1/2[180-(180n-360)/n]

Not sure how to deal with second bracket but here goes: (90 +(90n-180)/n)

so that leaves: (180n-360)/n - 90 + (90n - 180/n)


Hmmm......think I'm losing the plot now....

You're doing fine so far. Just keep going! (Well, I see one error, in that you moved a parenthesis in the last line, writing (90n - 180/n) instead of (90n - 180)/n ).

Or, if you'd rather start over, you might note that we usually simplify from the inside out, just the way we evaluate. So before you move on from (180n-360)/n, you might first distribute that division: 180n/n - 360/n = 180 - 360/n. Removing inner parentheses first prevents some common errors, and makes things look simpler faster.

If you think about it, this formula I just got, 180 - 360/n, is exactly what I described: the internal angle is the supplement of the external angle. Starting with this formula (with no parentheses in it) makes the next step simpler: [180 - 360/n] - [180 - (180 - 360/n)]/2. Now eliminate the innermost parentheses, and so on.
 
Thanks Dr. Peterson,

I'll try to continue distributing the division to get: 180-360/n - 90 +90 - 180/n

= 180 - 540/n


Hmmm...that's different to the answer in the book...gone wrong somewhere.
 
Thanks Dr. Peterson,

I'll try to continue distributing the division to get: 180-360/n - 90 +90 - 180/n

= 180 - 540/n


Hmmm...that's different to the answer in the book...gone wrong somewhere.

That's what I got. What does the book say?

Let's check it. For a square, n=4, angle ACD is 45°. Your formula gives 180 - 540/4 = 180 - 135 = 45°.

How about a hexagon, n=6. Sketching it, I see that ACD = 90°. The formula gives 180 - 540/6 = 180 - 90 = 90°.

Looks good to me! I'm guessing that the book's answer is equivalent, but written differently. For example, it could be written with 180 and division by n factored out, as 180(n-3)/n. Is that it? If so, then you aren't wrong!
 
That's what I got. What does the book say?

Let's check it. For a square, n=4, angle ACD is 45°. Your formula gives 180 - 540/4 = 180 - 135 = 45°.

How about a hexagon, n=6. Sketching it, I see that ACD = 90°. The formula gives 180 - 540/6 = 180 - 90 = 90°.

Looks good to me! I'm guessing that the book's answer is equivalent, but written differently. For example, it could be written with 180 and division by n factored out, as 180(n-3)/n. Is that it? If so, then you aren't wrong!


The answer in the book is: Angle ACD= (270(n-2))/n - 90 which I don't understand. Just to clarify it is written as:


formula.jpg
 
The answer in the book is: Angle ACD= (270(n-2))/n - 90 which I don't understand. Just to clarify it is written as:

I have no idea why they end up with that form; it must come from some different approach.

But when your book gives an apparently different answer from yours, you just have to try to transform them into the same form. Since our form is simpler, let's simplify theirs:

(270(n-2))/n - 90 = (270n-540)/n - 90 = 270 - 540/n - 90 = 180 - 540/n

And that is our answer! I'd say ours is better than theirs, since it is simpler. But both are correct.
 
I have no idea why they end up with that form; it must come from some different approach.

But when your book gives an apparently different answer from yours, you just have to try to transform them into the same form. Since our form is simpler, let's simplify theirs:

(270(n-2))/n - 90 = (270n-540)/n - 90 = 270 - 540/n - 90 = 180 - 540/n

And that is our answer! I'd say ours is better than theirs, since it is simpler. But both are correct.


That's great -thanks for help-I often lack confidence even when I'm on the right track!
 
....or 180(n-3)/n

The form of this answer reminds me of another problem, which in turn suggests an entirely different way to solve the problem. Now that we have dealt with the issue of simplifying, and have an answer, this is worth looking at.

Circumscribe a circle about the regular n-gon. One way to derive the formula for the interior angle is that it is subtended by n-2 of the n equal arcs into which the vertices divide the circumference. The central angle for this combined arc is (n-2)*360/n, so the inscribed angle at a vertex is half of that: (n-2)*180/n.

Do the same thing for our angle ACD! This is subtended by all but 3 of the n arcs, so the same reasoning leads directly to the answer, (n-3)*180/n.

Clearly this, too, is not what the author did; but it takes a lot less simplification, doesn't it!
 
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