question_answer

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

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

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

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

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

Correct Answer: B

Solution :

[b] From figure, \[\angle BIC=2\times \angle BAC=120{}^\circ \] And      \[IB=IC\] \[\therefore \]      \[\angle IBD=\angle ICD\] \[=\frac{180{}^\circ -120{}^\circ }{2}=30{}^\circ \] Now,     \[\angle BID=90{}^\circ -30{}^\circ =60{}^\circ \]