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, \[\angle BAC=\frac{1}{2}\times \angle BOC\] \[[\because \angle OCB=\angle OBC=m{}^\circ ]\] \[\Rightarrow \] \[n{}^\circ =\frac{1}{2}\times (180{}^\circ -2m{}^\circ )=90{}^\circ -m{}^\circ \] \[\therefore \] \[m{}^\circ +n{}^\circ =90{}^\circ \]You need to login to perform this action.
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