Question

In the given figure
If $\angle \text{AOB}={45}^{\circ },\angle \text{BOC}={30}^{\circ }\text{and}\angle \text{AOD}={110}^{\circ }$
Find angle $\text{COD and BOD}$

Answer 1

Given $\angle \text{AOB}={45}^{\circ }$
$\angle \text{BOC}={30}^{\circ }\text{and}\angle \text{AOD}={110}^{\circ }$
To find :$\angle \text{COD and}\angle \text{BOD}$



$\angle \text{AOD}={110}^{\circ }$
$\Rightarrow \angle \text{(AOB+BOC+COD)}={110}^{\circ }$[$\text{from figure}$]
$\Rightarrow {45}^{\circ }+{30}^{\circ }+\angle {\text{COD}}^{\circ }={110}^{\circ }$ [Given]
$\Rightarrow \angle \text{COD}={110}^{\circ }-{75}^{\circ }$
$\Rightarrow \angle \text{COD}={35}^{\circ }$

$\angle \text{BOD}=\angle \text{(BOC+COD)}$[$\text{from figure}$]
$\Rightarrow \angle \text{BOD}={30}^{\circ }+{35}^{\circ }={65}^{\circ }$

Hence $\angle \text{COD}={35}^{\circ }\text{and}\angle \text{BOD}={65}^{\circ }$