Question

In the given figure, find $x.$ Further find $\angle BOC,\angle COD$ and $\angle AOD.$

Answer 1

Given: $\angle AOD=(x+10{)}^{\circ },\angle COD=x^{\circ }$

and $\angle BOC=(x+20{)}^{\circ }$

Clearly, $\angle AOD,\angle BOC$ and $\angle COD$ form a linear pair

$\Rightarrow \angle AOD+\angle BOC+\angle COD={180}^{\circ }$

$\Rightarrow (x+10{)}^{\circ }+x^{\circ }+(x+20{)}^{\circ }=180$

$\Rightarrow 3x^{\circ }+{30}^{\circ }={180}^{\circ }$

$\Rightarrow 3x^{\circ }={180}^{\circ }–{30}^{\circ }$

$\Rightarrow x=\frac{{150}^{\circ }}{3}$

$\Rightarrow x={50}^{\circ }$Now,

$\angle AOD=(x+10{)}^{\circ }$

$={50}^{\circ }+{10}^{\circ }={60}^{\circ }$

$\angle COD=x^{\circ }={50}^{\circ }$

$\angle BOC=(x+20{)}^{\circ }$

$={50}^{\circ }+{20}^{\circ }={70}^{\circ }$

Hence, $x={50}^{\circ },\angle AOD={60}^{\circ },\angle COD={50}^{\circ }$ and $\angle BOC={70}^{\circ }.$