+0  
 
0
34
1
avatar

In the given figure O is the centre of the circle, BC and CD are two equal chords and angle OBC = 70 degrees. You have to find sum of angle BAC + angle BFD + angle CED (in degrees).

 

 Jan 3, 2021
 #1
avatar+114171 
+1

Connect  OC

 

Then   triangle BOC is isoceles  with   OC   =   OB,,  so angles OBC  and  OCB are equal

And central angle COB   =180  -2(70)   =  40°

 

So angle BAC  is an inscribed angle  intersecting the same arc as angle COB...so it measures (1/2) 40  = 20°

 

Note  that  arc   BCD  = 2 (40)  = 80°

And  angle BFD  =  (1/2)arc BCD   =  40°

 

And angle CED  =  angle BAC  =  20°

 

So.....BAC + BCD  + CED   =   20  +  40   +  20   =   80°

 

 

cool cool cool

 Jan 3, 2021

40 Online Users

avatar
avatar
avatar
avatar