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


ANSWER

$${ 16 }^{ 0 }$$


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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