Question

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

Answer 1

In the figure,
AOB is a straight line
$\therefore \angle AOD+\angle DOB={180}^{\circ }$ (Linear pair)
$\Rightarrow \angle AOD+\angle DOC+\angle COB={180}^{\circ }\Rightarrow x+{10}^{\circ }+x+x+20={180}^{\circ }\Rightarrow 3x+{30}^{\circ }={180}^{\circ }\Rightarrow 3x={180}^{\circ }-{30}^{\circ }={150}^{\circ }\Rightarrow x=\frac{{150}^{\circ }}{3}={50}^{\circ }\therefore x={50}^{\circ }$
Now $\angle BOC=x+{20}^{\circ }={50}^{\circ }+{20}^{\circ }={70}^{\circ }\angle COD=x={50}^{\circ }$
and $\angle AOD=x+{10}^{\circ }={50}^{\circ }+{10}^{\circ }={60}^{\circ }$