question_answer

In the given figure, 0 is the centre of in circle inscribed in\[\Delta \text{ABC}\]. Find \[\angle \text{BOC}\]if\[\angle \text{BAC}=\text{4}0{}^\circ \].

A)  \[~\text{1}00{}^\circ \]    

B)                         \[\text{12}0{}^\circ \]

C)  \[~\text{9}0{}^\circ \]                  

D)        \[~\text{11}0{}^\circ \]

Correct Answer: D

Solution :

 AO is joined. Since the circle is the in circle for\[\Rightarrow \], AO, BO and CO are the angle bisectors of \[{{6}^{x}}\] and \[\therefore \]respectively. \[{{6}^{x}}\] \[{{x}^{3}}\] In \['x'\] From the figure, \[\text{(25}\times \text{7)cm}\] \[~(\text{2}\times \text{52}\times \text{73) cm}\]