question_answer

In the figure shown, O is the centre of the circle, \[\angle AOB=25{}^\circ \]and \[\angle BOC\]then \[\angle BOC\]is equal to

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, \[\angle BAC=\angle BAO+\angle OAC\] \[=25+30=55{}^\circ \] \[\angle BOC=2\times \angle BAC{}^\circ \] \[=2\times 55=110{}^\circ \]