question_answer

In the following figure 'O' is the centre of circle and \[\angle BAC=n{}^\circ \],\[\angle OCB=m{}^\circ .\] Then,

A)  \[m+n=90{}^\circ \]   

B)  \[m+n=180{}^\circ \]

C)  \[m+n=120{}^\circ \] 

D)  \[m+n=150{}^\circ \]

Correct Answer: A

Solution :

\[\angle BAC=n{}^\circ \]and \[\angle OCB=m{}^\circ \] Then, \[\angle BAC=\frac{1}{2}\times \angle BOC\] \[[\because \angle OCB=\angle OBC=m{}^\circ ]\]             \[\Rightarrow \]   \[n{}^\circ =\frac{1}{2}\times (180{}^\circ -2m{}^\circ )=90{}^\circ -m{}^\circ \]             \[\therefore \] \[m{}^\circ +n{}^\circ =90{}^\circ \]