Mathematics

In the figure given $$ L_1$$ and $$L_2$$ are parallel. The sum of the angles $$\alpha, \beta, \gamma$$ marked in the diagram is: 


ANSWER

$$360^0$$


SOLUTION
Extend the line AO to $$OA^{'} $$

$$\Rightarrow \angle OA^{'} B=180-\alpha $$

$$\Rightarrow \angle A^{'} OB=180-\beta $$

$$\Rightarrow \angle A^{'} OB=180-\gamma $$

As $$OA^{'} B$$ is a triangle, sum of its angles =180 degrees 

$$\Rightarrow 540-(\alpha +\beta +\gamma) =180$$

$$\Rightarrow \alpha +\beta +\gamma =360$$
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