Question

In the given figure:
(i) If $\angle AOB={45}^{\circ },\angle BOC={30}^{\circ },\angle AOD={110}^{\circ }$ find the angles $COD$ and $BOD$.

Answer 1

Given:$\angle AOB={45}^{\circ }$

$\angle BOC={30}^{\circ }$ and $\angle AOD={110}^{\circ }$

To find :$\angle COD\text{and}\angle BOD$

$\angle AOD={110}^{\circ }$

$\angle AOB+\angle BOC+\angle COD={110}^{\circ }$[from figure]

${45}^{\circ }+{30}^{\circ }+\angle COD={110}^{\circ }$ [given]

$\angle COD={110}^{\circ }-{75}^{\circ }$

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

$\therefore$$\angle BOD=\angle BOC+\angle COD$

$\Rightarrow$ $\angle BOD={30}^{\circ }+{35}^{\circ }={65}^{\circ }$

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