Mathematics

# In the given figure, if $\angle BOC=7x+{ 20 }^{ 0 }\quad and\quad \angle COA=3x$, then the value of x for which AOB becomes a straight line is

${ 16 }^{ 0 }$

##### SOLUTION
For $\angle AOB$ to be a straight line, it should be equal to $180^{\circ}$

$\Rightarrow 7x+20^{\circ}+3x=180^{\circ}$

$\Rightarrow 10x=180^{\circ}-20^{\circ}$

$\Rightarrow 10x=160^{\circ}$

$\therefore x=16^{\circ}$

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Single Correct Medium Published on 09th 09, 2020
Questions 120418
Subjects 10
Chapters 88
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