Question

In the given alongside :
If $\angle AOB={45}^o,\angle BOC={30}^o$ and $\angle AOD={110}^o$, find angles $COD$ and $BOD$.

Answer 1

REF. Image.

Given $\angle AOB={45}^{\circ },\angle BOC={30}^{\circ },\angle AOD={110}^{\circ }$

we have to find $\angle COD$ & $\angle BOD$

i) $\angle COD=\angle AOD-(\angle AOB+\angle BOC)$

$=110-(45+30)=110-75={35}^{\circ }$

$\angle COD={35}^{\circ }$

ii) $\angle BOD=\angle AOD-\angle AOB$

$=110-{45}^{\circ }={65}^{\circ }$

$\therefore \angle BOD={65}^{\circ }$