question_answer

From the circumcentre I of the\[\Delta ABC,\] perpendicular ID is drawn on BC. If\[\angle BAC=60{}^\circ ,\]then the value of \[\angle BID\] is.

[SSC (CGL) Mains 2012]

A) \[75{}^\circ \]                          

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

C) \[45{}^\circ \]                          

D) \[80{}^\circ \]

Correct Answer: B

Solution :

From figure,

\[\angle BIC=2\times \angle BAC=120{}^\circ \]

[\[\because \]Angle made by same chord in the center of the circle is double to angle at circumference of the circle]

and       \[IB=IC\]                       [\[\therefore \]radius of the circle]

\[\therefore \]\[\angle IBD=\angle ICD=\frac{180{}^\circ -120{}^\circ }{2}=30{}^\circ \]

Now,     \[\angle BID=90{}^\circ -30{}^\circ =60{}^\circ \]