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Stuck on this problem: In the figure, $O$ is the center of the circle, $\angle BAO = 30^\circ$ and $\angle BCO = 40^\circ$. Find the shaded angle $ \angle AOC $.

 

 Jan 29, 2021
 #1
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Connect BO  and  note  that BO = AO = CO   since they are all radii

 

Then   in triangle  ABO,  AO  = BO    so  angle BAO  = angle   OBA  = 30

Similarly, in triangle CBO ,  CO  = BO   so  angle  BCO  = angle  OBC  = 40

 

And angle ABC  =angle OBA  + angle OBC = 30 + 40  = 70

 

But  angle  ABC  is an inscribed  angle interceptin the same arc as central angle AOC

 

So     AOC  = 2 * ABC =  2 * 70    =   140°

 

 

cool cool cool

 Jan 29, 2021

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