Sum of Exterior Angles
Problem Statement
What is the sum of the exterior angles in a triangle? in a quadrilateral? in a pentagon?
What is true about the sum of the exterior angles in any convex polygon? Why?
Problem setup
I will measure the exterior angles of several different convex polygons (triangle, quadrilateral,
pentagon, and nonagon) to discover the sum of the exterior angles of each. Next I will measure
the exterior angles of a concave polygon to see if its sum holds the same relationship. Then I
will use algebra to prove why this relationship is true.
Plans to Solve/Investigate the Problem
I will use the Geometer’s Sketchpad to find the sum of the measures of the exterior angles of
several convex polygons, starting with the triangle and increasing the number of sides to include
the quadrilateral, pentagon, and nonagon. Then I will test the sum for a concave polygon and
compare it to the sum for the convex polygons. Then I will prove the formula algebraically.
Investigation/Exploration of the Problem
The sum of the exterior angles of a triangle is 360°.
mDAB+mFBC+mECA = 360.00
E
m ECA = 100.11
C
A
m DAB = 134.59
m FBC = 125.31
B
D
The sum of the exterior angles of a quadrilateral is 360°.
F
mHAB+mCBG+mFCD+mEDA = 360.00
E
D
mEDA = 94.24
F
mFCD = 94.60
C
mCBG = 59.22
A
B
mHAB = 111.94
H
The sum of the exterior angles of a pentagon is 360°.
mJAB+mGBC+mHCD+mIDE+mFEA = 360.00
I
mIDE = 59.23
D
mHCD = 69.86
E
H
F
mFEA = 83.09
C
mGBC = 71.29
A
mJAB = 76.52
J
The sum of the exterior angles of a nonagon is 360°.
B
G
G
mNAB+mOBC+mPCD+mQDE+mREF+mJFG+mKGH+mLHI+mMIA = 360.00
J
mJFG = 47.41
R
F
G
Q
mREF = 35.23 E
mKGH = 36.49
K
H
mQDE = 40.09
D
mLHI = 18.69
L
I
P
mMIA = 21.46
mPCD = 48.44
C
M
mNAB = 72.26
mOBC = 39.93
A
B
O
N
Supported by the evidence above, we make the conjecture that the sum of the exterior angles of a
convex polygon is 360°.
Now let’s investigate the sum of the exterior angles of a concave polygon. I will use a pentagon.
F
mFAE+mABG+mBCH+mCDJ+mAED = 540.00
mFAE = 136.10
A
mAED = 141.45
J
E
D
mDEK = 38.55
mABG = 66.75
G
K
B
mBCH = 73.49
H
C
mCDJ = 122.21
We see that the sum of the exterior angles of the above figure do not add to be 360°. Actually,
there are only four exterior angles, and the fifth (DEK ) does not meet the definition of an
exterior angle of a triangle, as it is interior to the polygon. If we were to say that AED were
the exterior angle (which it is not because it is not formed by an extension of the side of the
polygon, using the dashed ray FK), even then the sum of the exterior angles would not be 360°,
as illustrated by the drawing below:
mFAE+mABG+mBCH+mCDJ+mAED = 540.00
mFAE = 136.10
A
mAED = 141.45
J
E
D
mCDJ = 122.21
mDEK = 38.55
mABG = 66.75
G
K
B
mBCH = 73.49
C
H
Next I will prove algebraically that the sum of the exterior angles of a convex polygon is 360°.
We know the sum of straight angles at the vertices of a convex n-gon is n(180°).
We know the sum of the interior angles of a convex n-gon is 180°(n-2).
Therefore the sum of the exterior angles of a convex polygon could be represented by:
n(180°) - 180°(n-2).
Factoring the expression, we get: 180°[n – (n-2)].
Simplifying, we get 180°(n – n + 2) = 180°(2) = 360°.
Thus, the sum of the exterior angles of a convex polygon is 360°.
Author & Contact
Clare S. Deaver
bdeaver3@aol.com