Math, asked by tTilak, 1 year ago

central angle theorem ????with diagram.

Answers

Answered by AzzyLand
6

Hi my dear friend,

Answer: The central angle subtended by two points on a circle is twice the inscribed angle subtended by those points.



Step-by-step:



let there be a circle with center O . arc AB intends AOB at the center and ACB ar any point C on the remaining part of the circle .



TO PROVE :- /_ AOB = 2( /_ ACB)



CONSTRUCTION :- join CO and produce it to any point D



PROOF :-



OA = OC [radii of same circle ]


/_ OAC = /_ ACO


[angles opp to equal side's of a triangle are equal]



/_ AOD = /_OAC + /_ACO


[ext angles = sum of equal opp angles]


/_AOD = 2(/_ACO)-------------(1)


[/_OAC = /_ACO]



similarly,


/_ DOB = 2(/_OCB) -------------(2)



In fig (i) and (iii)



adding (1) And (2)


/_AOD + /_ DOB = 2(/_ACO) + 2(/_OCB)


/_AOD + /_ DOB = 2(/_ACO + /_OCB)


/_AOB = 2(/_ACB)



In fig (ii)



subtracting (1) from (2)


/_DOB - /_DOA = 2(/_OCB - /_ACO)


/_AOB = 2(/_ACB)



hence in all cases we see


/_AOB = 2(/_ACB)

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Answered by dhruvbadaya1
3

Answer: The central angle subtended by two points on a circle is twice the inscribed angle subtended by those points.


Step-by-step proof:


let there be a circle with center O . arc AB intends AOB at the center and ACB ar any point C on the remaining part of the circle .


TO PROVE :- /_ AOB = 2( /_ ACB)


CONSTRUCTION :- join CO and produce it to any point D


PROOF :-


OA = OC [radii of same circle ]

/_ OAC = /_ ACO

[angles opp to equal side's of a triangle are equal]


/_ AOD = /_OAC + /_ACO

[ext angles = sum of equal opp angles]

/_AOD = 2(/_ACO)-------------(1)

[/_OAC = /_ACO]


similarly,

/_ DOB = 2(/_OCB) -------------(2)


In fig (i) and (iii)


adding (1) And (2)

/_AOD + /_ DOB = 2(/_ACO) + 2(/_OCB)

/_AOD + /_ DOB = 2(/_ACO + /_OCB)

/_AOB = 2(/_ACB)


In fig (ii)


subtracting (1) from (2)

/_DOB - /_DOA = 2(/_OCB - /_ACO)

/_AOB = 2(/_ACB)


hence in all cases we see

/_AOB = 2(/_ACB)

(proved)



Attachments:

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