Question: 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
Options:
A) $ 40{}^\circ $
B) $ 60{}^\circ $
C) $ 70{}^\circ $
D) $ 90{}^\circ $
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Answer:
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 $
BETA
