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:

$360^0$

##### 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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Single Correct Medium Published on 09th 09, 2020
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