Question

In the figure , O is the centre of the circle , BC=OB $\angle ACD=y$ and $\angle AOD=x$ . Then

• A. $x=y$
• B. $x=3y$
• C. $x=4y$
• D. $x=2y$

Answer 1

The correct option is B $x=3y$

$\Rightarrow$ $O$ is the center of the circle. $AB$ is the chord produced to $C$ such that $OB=BC$. $CO$ produced intersects the circle in $D$.

$\angle OCB=y$ and $\angle AOD=x$.

In $△OBC,$

$OB=BC$ [ Given ]

$\therefore$ $\angle OCB=\angle BOC$ [ Equal sides have equal angles opposite to them ]

$\Rightarrow$ $\angle BOC=y$

$\angle OBA=\angle BOC+\angle OCB$ [ Exterior angle of a triangle is equal to sum of its interior opposite angles ]

$\therefore$ $\angle OBA=y+y=2y$ ------ ( 1 )

In $△AOB,$

$OA=OB$ [ Radius of the circle ]

$\therefore$ $\angle OBA=\angle OAB$ [ Equal sides have equal angles opposite to them ]

$\Rightarrow$ $\angle OAB=2y$ [ Using ( 1 ) ]

In $△AOC,$

$\angle AOD=\angle OAC+\angle OCA$ [ Exterior angle of a triangle is equal to sum of its interior opposite angles ]

$\therefore$ $x=2y+y$ [ $\angle OAC=\angle OAB$ ]

$\therefore$ $x=3y$