Question

in the figure angle BAD=40 then find angle BCD

Answer 1

In ∆OAB:-
OA=OB [Each equal to radius]
So,
Angle OAB= Angle OBA [Isosceles∆]
Now,
Angle OAB+Angle OBA+Angle AOB=180°
40°+40°+AOB=180°
AOB= 100°
Now,
Angle AOB= Angle COD [Vert.Opp.Ang.]
Angle COD= 100°
Now,
OC=OD [Each equal to radius]
Angle OCD= Angle ODC [Isosceles∆]
In ∆COD:-
Angle OCD+Angle ODC+Angle COD=180°
2Angle OCD= 180°-100°
Angle OCD= 40°

OR,
Angle OAB= Angle OCD
Since,
Angles in the same segment.

Hope this helps you!

Answer 2

heya..

here is you answer..