Question

In the given figure, O is the centre of the circle. A is any point on minor arc BC. Find the value of $\angle BAC-\angle OBC$.

Answer 1

Here OB = OC = radius
$\Rightarrow \Delta OBC$ is isosceles
$\therefore \angle OBC=\angle OCB=y^{\circ }$
Now in $\Delta OBC$, by angle sum property
$\angle OBC+\angle OCB+\angle BOC={180}^{\circ }$
$y+y+t={180}^{\circ }$
$2y+t={180}^{\circ }$
$2y+({360}^{\circ }-\angle z)={180}^{\circ }$ [$\angle z$ is the reflex $\angle$ of $\angle t$]
$2y+{360}^{\circ }-\angle z={180}^{\circ }$



Reflex $\angle z=2x$ [Angle subtended by an arc at the centre is twice the angle subtented by it at the circumference]
$\Rightarrow 2y+{360}^{\circ }-2x={180}^{\circ }\Rightarrow 2y-2x={180}^{\circ }-{360}^{\circ }\Rightarrow 2y-2x=-{180}^{\circ }\Rightarrow y-x=-{90}^{\circ }\Rightarrow x-y={90}^{\circ }$
Thus $\angle BAC-\angle OBC={90}^{\circ }$