Prove that the sum of all the angles formed on the same side of a line at a given point on the line is 180° .
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Heya !!
Given :- AOB is a straight line and rays OC , OD and OE stand on it , forming Angle AOC , Angle COD and DOE and And Angle EOB.
To prove :- Angle AOC + Angle COD + Angle EOB + Angle DOE = 180°
Proof :- Ray OC stands on line AB.
Therefore,
Angle AOC + Angle COB = 180°
=> Angle AOC + ( Angle COD + Angle DOE + Angle EOB ) = 180° [ Since Angle COB = Angle COD + Angle DOE + Angle EOB )
=> Angle AOC + Angle COD + Angle DOE + Angle EOB = 180°
Hence,
The sum of all the angles formed on the same side of a line AB at a point O on it is 180°.
Given :- AOB is a straight line and rays OC , OD and OE stand on it , forming Angle AOC , Angle COD and DOE and And Angle EOB.
To prove :- Angle AOC + Angle COD + Angle EOB + Angle DOE = 180°
Proof :- Ray OC stands on line AB.
Therefore,
Angle AOC + Angle COB = 180°
=> Angle AOC + ( Angle COD + Angle DOE + Angle EOB ) = 180° [ Since Angle COB = Angle COD + Angle DOE + Angle EOB )
=> Angle AOC + Angle COD + Angle DOE + Angle EOB = 180°
Hence,
The sum of all the angles formed on the same side of a line AB at a point O on it is 180°.
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Refer to attached Image
Given = AOM is a straight line and ray OP , OQ and OR stand on it
To Prove = ∠AOP + ∠POQ + ∠QOR + ∠ROM = 180
Proof = Ray OP stands on line AM
∠AOP + ∠POM = 180
∠AOP + ∠POQ + ∠QOR + ∠ROM = 180 [Explanation ∠POQ + ∠QOR + ∠ROM = ∠POM]
∠AOP + ∠POQ + ∠QOR + ∠ROM = 180
Therefore
Sum of all angles formed on same side of line AM at a point O on it is 180
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