Mathematics

In the given figure , AOC is the diameter of the circle , with centre O. 
if arc AXB is half of arc BYC , find $$\angle BOC$$. 


SOLUTION
Given $$arc(A\times B)=\dfrac{1}{2}arc B\times C$$
$$\Rightarrow \dfrac{180^o-\theta}{360}\times 2\pi r=\dfrac{1}{2}\times \dfrac{\theta}{360^o}\times 2\pi r$$
$$\Rightarrow 360^o-2\theta =\theta$$
$$\Rightarrow \theta =120^o$$
$$\therefore \angle BOC=120^o$$.
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Subjective Medium Published on 09th 09, 2020
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