Question

Question 93

In th egiven figure, OB is perpendicular to OA and$\angle BOC={49}^{\circ }.$ Find $\angle AOD.$

Answer 1

From the given figure, we have
$\angle DOB+\angle BOC={180}^{\circ }\left[linearpair\right]\Rightarrow \angle DOB+{49}^{\circ }={180}^{\circ }\Rightarrow \angle DOB={180}^{\circ }-{49}^{\circ }={131}^{\circ }$
Now, $\angle DOB+\angle BOA+\angle AOB={360}^{\circ }[\therefore Sumofallanglesaroundapointis{360}^{\circ }]\Rightarrow {131}^{\circ }+{90}^{\circ }+\angle AOD={360}^{\circ }[\therefore \angle DOB={131}^{\circ },\angle BOA={90}^{\circ }]\Rightarrow {221}^{\circ }+\angle AOD={360}^{\circ }\Rightarrow \angle AOD={360}^{\circ }-{221}^{\circ }={139}^{\circ }$