how to proof that angle AEB =60 degree
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join OC, OD and BC
OC=OD=CD= RADIUS
THEREFORE TRIANGLE ABC IS EQUILATERAL
|_ COD=60°
NOW,
|_ CBD= ½|_ COD
|_CBD=30°
SINCE |_ ACB is an angle in a semi circle
therefore,|_ BCE=180°- |_ACB = 180°-90°
Thus, in triangle BCE
|_ BCE=90° AND ANGLE |_ CBD=30°
THEREFORE,
|_ BCE+ |_ CBE +|_ CEB= 180°
90°+30°+|_ CBE= 180°
ANGLE CEB = 60°
|_ AEB= 60°
HOPE IT HELP..:-D
OC=OD=CD= RADIUS
THEREFORE TRIANGLE ABC IS EQUILATERAL
|_ COD=60°
NOW,
|_ CBD= ½|_ COD
|_CBD=30°
SINCE |_ ACB is an angle in a semi circle
therefore,|_ BCE=180°- |_ACB = 180°-90°
Thus, in triangle BCE
|_ BCE=90° AND ANGLE |_ CBD=30°
THEREFORE,
|_ BCE+ |_ CBE +|_ CEB= 180°
90°+30°+|_ CBE= 180°
ANGLE CEB = 60°
|_ AEB= 60°
HOPE IT HELP..:-D
saxenayashi890:
mark it brainliest plzzz....
I have also prooved it in one of my maths questions u can check my profile's maths question you will find it with pic too.
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