in the figure if angle aob = 125 degree then angle COD is equal to
Attachments:
star1222:
hii
Answers
Answered by
27
Thank you for asking this question. Here is your answer:
The quadrilateral angle's circle meet at the point of the supplementary angles and that is at the center.
ABCD will meet at a point where it won't cut it but it will just pass it, and that would be the center at O.
∠AOB + ∠COD = 180°
125° + ∠COD = 180°
∠COD = 180° - 125°
∠COD = 55°
If there is any confusion please leave a comment below.
Is there any such theorem.... bcoz I m in class 10 and I haven't studied such theorems...... can u explain me in brief?
same problem here
with me
Same
I can't understand
what we can do now , and this question is most probably for board exam.
Ya.....i got the answer..... this theorem isn't in cbse 10th course... yet we can prove the answer.... I have proved that the opposite angles are supplementary at the centre..... hence the answer will be 55°....thanks to my teacher who helped me...
Answered by
3
Step-by-step explanation:
Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.
Hence,
∠COD=180°−125°
=55 °
Similar questions