Asked by ashly
prove that the exterior angle of polygon is 360
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MathMate
From the centre of a polygon of n-sides (n-gon), draw rays towards the n-vertices thus forming n triangles.
The sum of internal angles of the n triangles equals 180n°. This sum also equals the sum of the n interior angles plus the angles around the central point, 360°. Call this sum A.
Extend each side of the n-gon to create an exterior angle. The sum of the exterior angle and the corresponding interior angle equals 180°. So the sum of all the exterior angles and interior angles of the n-gon equals 180n°. Call this sum B.
Since both equal 180n°,
sum A = sum B, or
360°+∑interior angles = ∑exterior angles + ∑ interior angles
Cancel out ∑interior angles, we get
∑exterior angles = 360°.
The sum of internal angles of the n triangles equals 180n°. This sum also equals the sum of the n interior angles plus the angles around the central point, 360°. Call this sum A.
Extend each side of the n-gon to create an exterior angle. The sum of the exterior angle and the corresponding interior angle equals 180°. So the sum of all the exterior angles and interior angles of the n-gon equals 180n°. Call this sum B.
Since both equal 180n°,
sum A = sum B, or
360°+∑interior angles = ∑exterior angles + ∑ interior angles
Cancel out ∑interior angles, we get
∑exterior angles = 360°.
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