A) \[85{}^\circ \]
B) \[150{}^\circ \]
C) \[110{}^\circ \]
D) \[27\frac{1{}^\circ }{2}\]
Correct Answer: C
Solution :
[c] We know, \[\angle ABO=25{}^\circ \]and is equal to \[\angle OAC=25\] (OA = OB, so opposite angles are also equal) Similarly \[\angle AOC=30{}^\circ =\angle OAC\] Thus, \[\angle BAC=\angle BAO+\angle OAC\] \[=25+30=55{}^\circ \] \[\angle BOC=2\times \angle BAC{}^\circ \] \[=2\times 55=110{}^\circ \] |
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