In the fig. angle ADC = 130° and chord BC= chord BE. Find angle CBE.
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Sagittarius29:
please be fast
I need it urgently
Is there any more specifications about the question.......like if o is the centre or AB is the diameter?
no, the centre is not goven in this question
neither the diameter
*given
Okayy
Answers
Answered by
499
in triangle BCO and B0E
CO = OE (radii of same circle)
bo = bo ( common)
BC = BE(given)
so by SSS congruence rule
triangle BCO is congruent to triangle BOE
BY CPCT
angle CBO = angle OBE ( eqn 1)
since abcd is a cyclic quad.
therefore angle D + angle B = 180
130 + angle B = 180
angle B = 50 degree
angle CBO + angle OBE = angle CBE
BUt angle cbo = angle OBE
therefore 50 + 50 = angle CBE
angle CBE = 100 degree
donot forget to make it brainlist answer.
CO = OE (radii of same circle)
bo = bo ( common)
BC = BE(given)
so by SSS congruence rule
triangle BCO is congruent to triangle BOE
BY CPCT
angle CBO = angle OBE ( eqn 1)
since abcd is a cyclic quad.
therefore angle D + angle B = 180
130 + angle B = 180
angle B = 50 degree
angle CBO + angle OBE = angle CBE
BUt angle cbo = angle OBE
therefore 50 + 50 = angle CBE
angle CBE = 100 degree
donot forget to make it brainlist answer.
We have to solve this question using o as a centre.....
is it correct?
I had exempler of 9th the solution has been made taking o as the centre
konsa wala question?
I saw it on the website, yes you are right Eshita2416 and Deepak70
The solution is done using O as the centre
Sorry for the inconvenience actually in my book it i not specified, so I thought....
Please don't mind
*is
np..
Answered by
164
ABCD is cyclic quadrilateral.
Angle ADC + Angle CBA= 180
See the photo for complete answer.
Thanks..
Angle ADC + Angle CBA= 180
See the photo for complete answer.
Thanks..
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