If the angle of a quadrilateral are $x^{0,}{(x-10)}^0,{(x+30)}^{0 }and 2x^0$, then the greatest angle is:

 

Concept: The sum of four interior angles of a quadrilateral is ${360}^0$.

The given angles of a quadrilateral are  

$x^{0,}{(x-10)}^0,{(x+30)}^{0 }and 2x^0$

$x^0 {+(x-10)}^0+{ (x+30)}^{0 }+ 2x^0={360}^0$

$\Rightarrow 5x={360}^0-{20}^0={340}^0$

$\Rightarrow x=\frac{{340}^0}{5}$

Now, a measure of angles:

$x^0={68}^0$

${(x-10)}^0={(68-10)}^0={58}^0$

$(x+{30}^0)={(68+30)}^0={98}^0$ and $2x^0=\left({68}^0\right)={136}^0$

$\therefore$ The greatest angle is ${136}^0$

Hence, option 1 is correct.