question_answer

The circumcentre of a \[\Delta ABC\] is O. If  \[\angle BAC=85{}^\circ \]and \[\angle BCA=75{}^\circ ,\]then the value of\[\angle OAC\]is [SSC (10+2) 2012]

A) \[40{}^\circ \]                          

B) \[60{}^\circ \]

C) \[70{}^\circ \]  

D) \[90{}^\circ \]

Correct Answer: C

Solution :

\[\because \]\[\angle BAC=85{}^\circ \]

\[\therefore \]\[\angle BOC=2\times 85{}^\circ =170{}^\circ \]

[since, angle subtended by an arc at the centre of a circle is twice the angle subtended by the arc at any point on the remaining part of circle]

In \[\Delta BOC,\]\[OB=OC\]                   [radii of circle]

So, \[\angle OBC=\angle OCB\]

Now, \[\angle BOC+\angle OBC+\angle OCB=180{}^\circ \]

\[\Rightarrow \]   \[170{}^\circ +\angle OBC+\angle OBC=180{}^\circ \]

\[\Rightarrow \]               \[2\angle OBC=180{}^\circ -170{}^\circ \]

\[\Rightarrow \]               \[\angle OBC=\frac{10{}^\circ }{2}=5{}^\circ \]

Now, \[\angle OCB+\angle OCA=75{}^\circ \]

\[\angle OCA=75{}^\circ -5{}^\circ =70{}^\circ \]

In \[\Delta AOC,\]           \[OC=OA\]        [radii of circle]

so that   \[\angle OCA=\angle OAC=70{}^\circ \]