Prove that a regular polygon has five sides, if its exterior angle is two third of its interior angle
Answers
Answered by
47
Given:
Exterior ∠= 2/3 of interior∠
Exterior ∠ + interior adjacent ∠ + 180°
Exterior ∠ = 360°/ n ( n= no. of sides)
Each interior ∠ = 180 - exterior ∠
Each interior ∠ = 180 - 360° /n
A.T.Q
Exterior ∠= 2/3 of interior∠
360°/ n = 2/3 ( 180° - 360° /n)
360°/ n = 2/3 ×180° - 2/3 × 360°/n
360°/ n = 120° - 240° /n
360°/ n +240°/n = 120°
600 °/n =120°
n= 600°/120°
n= 5
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Hence polygon has 5 sides....
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Hope this will help you.....
Exterior ∠= 2/3 of interior∠
Exterior ∠ + interior adjacent ∠ + 180°
Exterior ∠ = 360°/ n ( n= no. of sides)
Each interior ∠ = 180 - exterior ∠
Each interior ∠ = 180 - 360° /n
A.T.Q
Exterior ∠= 2/3 of interior∠
360°/ n = 2/3 ( 180° - 360° /n)
360°/ n = 2/3 ×180° - 2/3 × 360°/n
360°/ n = 120° - 240° /n
360°/ n +240°/n = 120°
600 °/n =120°
n= 600°/120°
n= 5
----------------------------------------------------------------------------------------------------
Hence polygon has 5 sides....
---------------------------------------------------------------------------------------------------
Hope this will help you.....
Answered by
0
Answer:
5 sides
Step-by-step explanation:
A.T.Q ( According To The Question)
Exterior ∠= 2/3 of interior∠
360°/ n = 2/3 ( 180° - 360° /n)
360°/ n = 2/3 ×180° - 2/3 × 360°/n
360°/ n = 120° - 240° /n
360°/ n +240°/n = 120°
600 °/n =120°
n= 600°/120°
n= 5
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