Prove that, for any polygon, taking all pair of adjacent angles, subtracting 180 from their sum, and adding...
$begingroup$
Take any simple polygon. Extend all sides in both directions. Note the angles where these sides meet, if at all. What is the sum of these angles?
For example, consider the figure formed from a regular hexagon, the Star of David:
Given that the angles of a regular hexagon are each $120$ degrees, it’s easy to calculate that the angles in question are $60$ degrees each, since the base angles of the triangle are supplementary to the angles of the hexagon, and the angles of a triangle must add to $180$.
Therefore, since there are six of these, they add up to $360$ degrees in total.
Exchanging the $6$ for $n$ (for general case), this can be written as
$$nleft(180-2left(180-frac{180(n-2)}{n}right)right)$$
Distributing and simplifying:
$$180n-2nleft(180-frac{180(n-2)}{n}right)$$
$$180n-360n+360n-720$$
$$180n-720$$
$$180(n-4)$$
Another way of wording this is that this gives a measure of how far any two adjacent angles in a polygon are from forming parallel lines; that is, how far their angles are from $90$ degrees, with the result being positive if they’re slanted inward, and negative if they’re slanted outward. You can calculate this by taking each angle in the pair, subtracting $90$ from them, and adding the results together.
Mathematically, it’s the same as above, but it’s conceptually very different. The benefit of thinking of it this way is that this justifies a square giving an output of $0$ and an equilateral triangle giving an output of $-180$. And...well, when I discuss crazier cases later, thinking of the angles this way will make more sense, especially when they don’t exist.
This proof hinges on the original polygon being regular, however. The formula is derived from multiplying the angle by the number of triangles, and that’s only true if there are that many triangles.
Consider the following pentagon, with its lines extended:
This pentagon has three right angles and two $135$-degree angles. By the same method as above, the four triangles formed can be shown to have points of $45$ degrees each. The bottom of the pentagon doesn’t form a triangle at all, but, more specifically, it forms two parallel lines; as this doesn’t stray from $90$ degrees at all, this gets us a $0$ for the bottom, and $180$ overall.
This is exactly what you’d get from the above formula: $180(5-4)$.
Consider the figure formed from a trapezoid:
Two pairs of angles yield parallel lines and therefore output $0$ each. One pair actually yields a triangle, with a third angle measuring $90$; that is to say, each of those angles on the trapezoid leans inward from a perfectly horizontal line by $45$ degrees. The fourth pair doesn’t form a triangle, but more importantly, the lines lean outward by $45$ degrees. These two effects cancel out, and overall, the figure nets a $0$ - no different than a square.
It’s not just irregular polygons - concave ones follow this pattern as well. It would be too hard to draw how the “triangles” work here, but consider a quadrilateral with angles $90, 30, 30, 210$. By the same logic as above - by subtracting $90$ from each angle in each adjacent pair - you still end up with a result of $0$!
I can justify $180(n-4)$ holding true by regular polygons. How do I prove that this is the case for all polygons?
geometry
edited Dec 20 '18 at 4:29
Micah
30.2k1364106
asked Dec 20 '18 at 3:45
DonielFDonielF
515515
$endgroup$
add a comment |
$begingroup$
Take any simple polygon. Extend all sides in both directions. Note the angles where these sides meet, if at all. What is the sum of these angles?
For example, consider the figure formed from a regular hexagon, the Star of David:
Given that the angles of a regular hexagon are each $120$ degrees, it’s easy to calculate that the angles in question are $60$ degrees each, since the base angles of the triangle are supplementary to the angles of the hexagon, and the angles of a triangle must add to $180$.
Therefore, since there are six of these, they add up to $360$ degrees in total.
Exchanging the $6$ for $n$ (for general case), this can be written as
$$nleft(180-2left(180-frac{180(n-2)}{n}right)right)$$
Distributing and simplifying:
$$180n-2nleft(180-frac{180(n-2)}{n}right)$$
$$180n-360n+360n-720$$
$$180n-720$$
$$180(n-4)$$
Another way of wording this is that this gives a measure of how far any two adjacent angles in a polygon are from forming parallel lines; that is, how far their angles are from $90$ degrees, with the result being positive if they’re slanted inward, and negative if they’re slanted outward. You can calculate this by taking each angle in the pair, subtracting $90$ from them, and adding the results together.
Mathematically, it’s the same as above, but it’s conceptually very different. The benefit of thinking of it this way is that this justifies a square giving an output of $0$ and an equilateral triangle giving an output of $-180$. And...well, when I discuss crazier cases later, thinking of the angles this way will make more sense, especially when they don’t exist.
This proof hinges on the original polygon being regular, however. The formula is derived from multiplying the angle by the number of triangles, and that’s only true if there are that many triangles.
Consider the following pentagon, with its lines extended:
This pentagon has three right angles and two $135$-degree angles. By the same method as above, the four triangles formed can be shown to have points of $45$ degrees each. The bottom of the pentagon doesn’t form a triangle at all, but, more specifically, it forms two parallel lines; as this doesn’t stray from $90$ degrees at all, this gets us a $0$ for the bottom, and $180$ overall.
This is exactly what you’d get from the above formula: $180(5-4)$.
Consider the figure formed from a trapezoid:
Two pairs of angles yield parallel lines and therefore output $0$ each. One pair actually yields a triangle, with a third angle measuring $90$; that is to say, each of those angles on the trapezoid leans inward from a perfectly horizontal line by $45$ degrees. The fourth pair doesn’t form a triangle, but more importantly, the lines lean outward by $45$ degrees. These two effects cancel out, and overall, the figure nets a $0$ - no different than a square.
It’s not just irregular polygons - concave ones follow this pattern as well. It would be too hard to draw how the “triangles” work here, but consider a quadrilateral with angles $90, 30, 30, 210$. By the same logic as above - by subtracting $90$ from each angle in each adjacent pair - you still end up with a result of $0$!
I can justify $180(n-4)$ holding true by regular polygons. How do I prove that this is the case for all polygons?
geometry
edited Dec 20 '18 at 4:29
Micah
30.2k1364106
asked Dec 20 '18 at 3:45
DonielFDonielF
515515
$endgroup$
add a comment |
$begingroup$
Take any simple polygon. Extend all sides in both directions. Note the angles where these sides meet, if at all. What is the sum of these angles?
For example, consider the figure formed from a regular hexagon, the Star of David:
Given that the angles of a regular hexagon are each $120$ degrees, it’s easy to calculate that the angles in question are $60$ degrees each, since the base angles of the triangle are supplementary to the angles of the hexagon, and the angles of a triangle must add to $180$.
Therefore, since there are six of these, they add up to $360$ degrees in total.
Exchanging the $6$ for $n$ (for general case), this can be written as
$$nleft(180-2left(180-frac{180(n-2)}{n}right)right)$$
Distributing and simplifying:
$$180n-2nleft(180-frac{180(n-2)}{n}right)$$
$$180n-360n+360n-720$$
$$180n-720$$
$$180(n-4)$$
Another way of wording this is that this gives a measure of how far any two adjacent angles in a polygon are from forming parallel lines; that is, how far their angles are from $90$ degrees, with the result being positive if they’re slanted inward, and negative if they’re slanted outward. You can calculate this by taking each angle in the pair, subtracting $90$ from them, and adding the results together.
Mathematically, it’s the same as above, but it’s conceptually very different. The benefit of thinking of it this way is that this justifies a square giving an output of $0$ and an equilateral triangle giving an output of $-180$. And...well, when I discuss crazier cases later, thinking of the angles this way will make more sense, especially when they don’t exist.
This proof hinges on the original polygon being regular, however. The formula is derived from multiplying the angle by the number of triangles, and that’s only true if there are that many triangles.
Consider the following pentagon, with its lines extended:
This pentagon has three right angles and two $135$-degree angles. By the same method as above, the four triangles formed can be shown to have points of $45$ degrees each. The bottom of the pentagon doesn’t form a triangle at all, but, more specifically, it forms two parallel lines; as this doesn’t stray from $90$ degrees at all, this gets us a $0$ for the bottom, and $180$ overall.
This is exactly what you’d get from the above formula: $180(5-4)$.
Consider the figure formed from a trapezoid:
Two pairs of angles yield parallel lines and therefore output $0$ each. One pair actually yields a triangle, with a third angle measuring $90$; that is to say, each of those angles on the trapezoid leans inward from a perfectly horizontal line by $45$ degrees. The fourth pair doesn’t form a triangle, but more importantly, the lines lean outward by $45$ degrees. These two effects cancel out, and overall, the figure nets a $0$ - no different than a square.
It’s not just irregular polygons - concave ones follow this pattern as well. It would be too hard to draw how the “triangles” work here, but consider a quadrilateral with angles $90, 30, 30, 210$. By the same logic as above - by subtracting $90$ from each angle in each adjacent pair - you still end up with a result of $0$!
I can justify $180(n-4)$ holding true by regular polygons. How do I prove that this is the case for all polygons?
geometry
edited Dec 20 '18 at 4:29
Micah
30.2k1364106
asked Dec 20 '18 at 3:45
DonielFDonielF
515515
$endgroup$
Take any simple polygon. Extend all sides in both directions. Note the angles where these sides meet, if at all. What is the sum of these angles?
For example, consider the figure formed from a regular hexagon, the Star of David:
Given that the angles of a regular hexagon are each $120$ degrees, it’s easy to calculate that the angles in question are $60$ degrees each, since the base angles of the triangle are supplementary to the angles of the hexagon, and the angles of a triangle must add to $180$.
Therefore, since there are six of these, they add up to $360$ degrees in total.
Exchanging the $6$ for $n$ (for general case), this can be written as
$$nleft(180-2left(180-frac{180(n-2)}{n}right)right)$$
Distributing and simplifying:
$$180n-2nleft(180-frac{180(n-2)}{n}right)$$
$$180n-360n+360n-720$$
$$180n-720$$
$$180(n-4)$$
Another way of wording this is that this gives a measure of how far any two adjacent angles in a polygon are from forming parallel lines; that is, how far their angles are from $90$ degrees, with the result being positive if they’re slanted inward, and negative if they’re slanted outward. You can calculate this by taking each angle in the pair, subtracting $90$ from them, and adding the results together.
Mathematically, it’s the same as above, but it’s conceptually very different. The benefit of thinking of it this way is that this justifies a square giving an output of $0$ and an equilateral triangle giving an output of $-180$. And...well, when I discuss crazier cases later, thinking of the angles this way will make more sense, especially when they don’t exist.
This proof hinges on the original polygon being regular, however. The formula is derived from multiplying the angle by the number of triangles, and that’s only true if there are that many triangles.
Consider the following pentagon, with its lines extended:
This pentagon has three right angles and two $135$-degree angles. By the same method as above, the four triangles formed can be shown to have points of $45$ degrees each. The bottom of the pentagon doesn’t form a triangle at all, but, more specifically, it forms two parallel lines; as this doesn’t stray from $90$ degrees at all, this gets us a $0$ for the bottom, and $180$ overall.
This is exactly what you’d get from the above formula: $180(5-4)$.
Consider the figure formed from a trapezoid:
Two pairs of angles yield parallel lines and therefore output $0$ each. One pair actually yields a triangle, with a third angle measuring $90$; that is to say, each of those angles on the trapezoid leans inward from a perfectly horizontal line by $45$ degrees. The fourth pair doesn’t form a triangle, but more importantly, the lines lean outward by $45$ degrees. These two effects cancel out, and overall, the figure nets a $0$ - no different than a square.
It’s not just irregular polygons - concave ones follow this pattern as well. It would be too hard to draw how the “triangles” work here, but consider a quadrilateral with angles $90, 30, 30, 210$. By the same logic as above - by subtracting $90$ from each angle in each adjacent pair - you still end up with a result of $0$!
I can justify $180(n-4)$ holding true by regular polygons. How do I prove that this is the case for all polygons?
geometry
geometry
edited Dec 20 '18 at 4:29
Micah
30.2k1364106
asked Dec 20 '18 at 3:45
DonielFDonielF
515515
edited Dec 20 '18 at 4:29
Micah
30.2k1364106
asked Dec 20 '18 at 3:45
DonielFDonielF
515515
edited Dec 20 '18 at 4:29
Micah
30.2k1364106
edited Dec 20 '18 at 4:29
Micah
30.2k1364106
edited Dec 20 '18 at 4:29
Micah
30.2k1364106
30.2k1364106
asked Dec 20 '18 at 3:45
DonielFDonielF
515515
asked Dec 20 '18 at 3:45
DonielFDonielF
515515
asked Dec 20 '18 at 3:45
DonielFDonielF
515515
515515
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For an n-gon,
considering a line
rotating around it
rotates 360.
This implies the standard result
that
the sum of the exterior angles
in the direction of the rotation
is 360
so the sum of the interior angles
is n*180-360 = 180(n-2).
Therefore
the sum of the exterior angles
on both sides of each vertex
is 720.
From your first diagram,
each angle of
the extended polygon
is 180 minus the
sum of the adjacent exterior angles,
so their sum is
180*n-720
=180(n-4).
answered Dec 20 '18 at 4:27
marty cohenmarty cohen
74.5k549129
$endgroup$
$begingroup$
I don’t follow how the first paragraph implies the second.
$endgroup$
– DonielF
Dec 20 '18 at 4:35
$begingroup$
There are two equal exterior angles adjacent to each interior angle. Since the sum of the exterior angles in, say, the clockwise direction is 360, the sum of them in both directions is 720.
$endgroup$
– marty cohen
Dec 20 '18 at 14:10
$begingroup$
I’m still not following. Where do you get the second rotation from? Maybe if you attached a diagram?
$endgroup$
– DonielF
Dec 20 '18 at 14:16
$begingroup$
I'm not sure whether my answer is the same as yours, just written differently. Can you confirm?
$endgroup$
– DonielF
Dec 24 '18 at 18:07
add a comment |
$begingroup$
I'm disappointed that I didn't see this earlier.
Another way to word my conclusion is that the formula I calculated is another way of saying that if you take every angle of the polygon and subtract $90$ from each, and you do this twice for each (one for each time you extend the line - one in each direction), you get $180(n-4)$. That is:
$$2sum_{i=1}^n(a_i-90)=180(n-4)$$
where $n$ is the number of angles in the polygon.
Since addition is commutative, the summation can be restated as
$$2left(sum_{i=1}^na_i-sum_{i=1}^n90right)$$
The first sigma is just adding up the angles of the polygon, which we know is $180(n-2)$, and the second one is the same as adding $90$ to itself $n$ times, so this can be rewritten as
$$2(180(n-2)-90n)$$
$$360n-720-180n$$
$$180n-720$$
$$180(n-4)$$
Q.E.D.
answered Dec 24 '18 at 18:07
DonielFDonielF
515515
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For an n-gon,
considering a line
rotating around it
rotates 360.
This implies the standard result
that
the sum of the exterior angles
in the direction of the rotation
is 360
so the sum of the interior angles
is n*180-360 = 180(n-2).
Therefore
the sum of the exterior angles
on both sides of each vertex
is 720.
From your first diagram,
each angle of
the extended polygon
is 180 minus the
sum of the adjacent exterior angles,
so their sum is
180*n-720
=180(n-4).
answered Dec 20 '18 at 4:27
marty cohenmarty cohen
74.5k549129
$endgroup$
$begingroup$
I don’t follow how the first paragraph implies the second.
$endgroup$
– DonielF
Dec 20 '18 at 4:35
$begingroup$
There are two equal exterior angles adjacent to each interior angle. Since the sum of the exterior angles in, say, the clockwise direction is 360, the sum of them in both directions is 720.
$endgroup$
– marty cohen
Dec 20 '18 at 14:10
$begingroup$
I’m still not following. Where do you get the second rotation from? Maybe if you attached a diagram?
$endgroup$
– DonielF
Dec 20 '18 at 14:16
$begingroup$
I'm not sure whether my answer is the same as yours, just written differently. Can you confirm?
$endgroup$
– DonielF
Dec 24 '18 at 18:07
add a comment |
$begingroup$
For an n-gon,
considering a line
rotating around it
rotates 360.
This implies the standard result
that
the sum of the exterior angles
in the direction of the rotation
is 360
so the sum of the interior angles
is n*180-360 = 180(n-2).
Therefore
the sum of the exterior angles
on both sides of each vertex
is 720.
From your first diagram,
each angle of
the extended polygon
is 180 minus the
sum of the adjacent exterior angles,
so their sum is
180*n-720
=180(n-4).
answered Dec 20 '18 at 4:27
marty cohenmarty cohen
74.5k549129
$endgroup$
$begingroup$
I don’t follow how the first paragraph implies the second.
$endgroup$
– DonielF
Dec 20 '18 at 4:35
$begingroup$
There are two equal exterior angles adjacent to each interior angle. Since the sum of the exterior angles in, say, the clockwise direction is 360, the sum of them in both directions is 720.
$endgroup$
– marty cohen
Dec 20 '18 at 14:10
$begingroup$
I’m still not following. Where do you get the second rotation from? Maybe if you attached a diagram?
$endgroup$
– DonielF
Dec 20 '18 at 14:16
$begingroup$
I'm not sure whether my answer is the same as yours, just written differently. Can you confirm?
$endgroup$
– DonielF
Dec 24 '18 at 18:07
add a comment |
$begingroup$
For an n-gon,
considering a line
rotating around it
rotates 360.
This implies the standard result
that
the sum of the exterior angles
in the direction of the rotation
is 360
so the sum of the interior angles
is n*180-360 = 180(n-2).
Therefore
the sum of the exterior angles
on both sides of each vertex
is 720.
From your first diagram,
each angle of
the extended polygon
is 180 minus the
sum of the adjacent exterior angles,
so their sum is
180*n-720
=180(n-4).
answered Dec 20 '18 at 4:27
marty cohenmarty cohen
74.5k549129
$endgroup$
For an n-gon,
considering a line
rotating around it
rotates 360.
This implies the standard result
that
the sum of the exterior angles
in the direction of the rotation
is 360
so the sum of the interior angles
is n*180-360 = 180(n-2).
Therefore
the sum of the exterior angles
on both sides of each vertex
is 720.
From your first diagram,
each angle of
the extended polygon
is 180 minus the
sum of the adjacent exterior angles,
so their sum is
180*n-720
=180(n-4).
answered Dec 20 '18 at 4:27
marty cohenmarty cohen
74.5k549129
answered Dec 20 '18 at 4:27
marty cohenmarty cohen
74.5k549129
answered Dec 20 '18 at 4:27
marty cohenmarty cohen
74.5k549129
answered Dec 20 '18 at 4:27
marty cohenmarty cohen
74.5k549129
74.5k549129
$begingroup$
I don’t follow how the first paragraph implies the second.
$endgroup$
– DonielF
Dec 20 '18 at 4:35
$begingroup$
There are two equal exterior angles adjacent to each interior angle. Since the sum of the exterior angles in, say, the clockwise direction is 360, the sum of them in both directions is 720.
$endgroup$
– marty cohen
Dec 20 '18 at 14:10
$begingroup$
I’m still not following. Where do you get the second rotation from? Maybe if you attached a diagram?
$endgroup$
– DonielF
Dec 20 '18 at 14:16
$begingroup$
I'm not sure whether my answer is the same as yours, just written differently. Can you confirm?
$endgroup$
– DonielF
Dec 24 '18 at 18:07
add a comment |
$begingroup$
I don’t follow how the first paragraph implies the second.
$endgroup$
– DonielF
Dec 20 '18 at 4:35
$begingroup$
There are two equal exterior angles adjacent to each interior angle. Since the sum of the exterior angles in, say, the clockwise direction is 360, the sum of them in both directions is 720.
$endgroup$
– marty cohen
Dec 20 '18 at 14:10
$begingroup$
I’m still not following. Where do you get the second rotation from? Maybe if you attached a diagram?
$endgroup$
– DonielF
Dec 20 '18 at 14:16
$begingroup$
I'm not sure whether my answer is the same as yours, just written differently. Can you confirm?
$endgroup$
– DonielF
Dec 24 '18 at 18:07
$begingroup$
I don’t follow how the first paragraph implies the second.
$endgroup$
– DonielF
Dec 20 '18 at 4:35
$begingroup$
I don’t follow how the first paragraph implies the second.
$endgroup$
– DonielF
Dec 20 '18 at 4:35
$begingroup$
There are two equal exterior angles adjacent to each interior angle. Since the sum of the exterior angles in, say, the clockwise direction is 360, the sum of them in both directions is 720.
$endgroup$
– marty cohen
Dec 20 '18 at 14:10
$begingroup$
There are two equal exterior angles adjacent to each interior angle. Since the sum of the exterior angles in, say, the clockwise direction is 360, the sum of them in both directions is 720.
$endgroup$
– marty cohen
Dec 20 '18 at 14:10
$begingroup$
I’m still not following. Where do you get the second rotation from? Maybe if you attached a diagram?
$endgroup$
– DonielF
Dec 20 '18 at 14:16
$begingroup$
I’m still not following. Where do you get the second rotation from? Maybe if you attached a diagram?
$endgroup$
– DonielF
Dec 20 '18 at 14:16
$begingroup$
I'm not sure whether my answer is the same as yours, just written differently. Can you confirm?
$endgroup$
– DonielF
Dec 24 '18 at 18:07
$begingroup$
I'm not sure whether my answer is the same as yours, just written differently. Can you confirm?
$endgroup$
– DonielF
Dec 24 '18 at 18:07
add a comment |
$begingroup$
I'm disappointed that I didn't see this earlier.
Another way to word my conclusion is that the formula I calculated is another way of saying that if you take every angle of the polygon and subtract $90$ from each, and you do this twice for each (one for each time you extend the line - one in each direction), you get $180(n-4)$. That is:
$$2sum_{i=1}^n(a_i-90)=180(n-4)$$
where $n$ is the number of angles in the polygon.
Since addition is commutative, the summation can be restated as
$$2left(sum_{i=1}^na_i-sum_{i=1}^n90right)$$
The first sigma is just adding up the angles of the polygon, which we know is $180(n-2)$, and the second one is the same as adding $90$ to itself $n$ times, so this can be rewritten as
$$2(180(n-2)-90n)$$
$$360n-720-180n$$
$$180n-720$$
$$180(n-4)$$
Q.E.D.
answered Dec 24 '18 at 18:07
DonielFDonielF
515515
$endgroup$
add a comment |
$begingroup$
I'm disappointed that I didn't see this earlier.
Another way to word my conclusion is that the formula I calculated is another way of saying that if you take every angle of the polygon and subtract $90$ from each, and you do this twice for each (one for each time you extend the line - one in each direction), you get $180(n-4)$. That is:
$$2sum_{i=1}^n(a_i-90)=180(n-4)$$
where $n$ is the number of angles in the polygon.
Since addition is commutative, the summation can be restated as
$$2left(sum_{i=1}^na_i-sum_{i=1}^n90right)$$
The first sigma is just adding up the angles of the polygon, which we know is $180(n-2)$, and the second one is the same as adding $90$ to itself $n$ times, so this can be rewritten as
$$2(180(n-2)-90n)$$
$$360n-720-180n$$
$$180n-720$$
$$180(n-4)$$
Q.E.D.
answered Dec 24 '18 at 18:07
DonielFDonielF
515515
$endgroup$
add a comment |
$begingroup$
I'm disappointed that I didn't see this earlier.
Another way to word my conclusion is that the formula I calculated is another way of saying that if you take every angle of the polygon and subtract $90$ from each, and you do this twice for each (one for each time you extend the line - one in each direction), you get $180(n-4)$. That is:
$$2sum_{i=1}^n(a_i-90)=180(n-4)$$
where $n$ is the number of angles in the polygon.
Since addition is commutative, the summation can be restated as
$$2left(sum_{i=1}^na_i-sum_{i=1}^n90right)$$
The first sigma is just adding up the angles of the polygon, which we know is $180(n-2)$, and the second one is the same as adding $90$ to itself $n$ times, so this can be rewritten as
$$2(180(n-2)-90n)$$
$$360n-720-180n$$
$$180n-720$$
$$180(n-4)$$
Q.E.D.
answered Dec 24 '18 at 18:07
DonielFDonielF
515515
$endgroup$
I'm disappointed that I didn't see this earlier.
Another way to word my conclusion is that the formula I calculated is another way of saying that if you take every angle of the polygon and subtract $90$ from each, and you do this twice for each (one for each time you extend the line - one in each direction), you get $180(n-4)$. That is:
$$2sum_{i=1}^n(a_i-90)=180(n-4)$$
where $n$ is the number of angles in the polygon.
Since addition is commutative, the summation can be restated as
$$2left(sum_{i=1}^na_i-sum_{i=1}^n90right)$$
The first sigma is just adding up the angles of the polygon, which we know is $180(n-2)$, and the second one is the same as adding $90$ to itself $n$ times, so this can be rewritten as
$$2(180(n-2)-90n)$$
$$360n-720-180n$$
$$180n-720$$
$$180(n-4)$$
Q.E.D.
answered Dec 24 '18 at 18:07
DonielFDonielF
515515
answered Dec 24 '18 at 18:07
DonielFDonielF
515515
answered Dec 24 '18 at 18:07
DonielFDonielF
515515
answered Dec 24 '18 at 18:07
DonielFDonielF
515515
515515
add a comment |
add a comment |
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