Question

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

• A. ${90}^{\circ }$
• B. ${120}^{\circ }$
• C. ${60}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is A ${90}^{\circ }$

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 $\angle z=2x$ [By degree measure theorem]
$\Rightarrow 2y+{360}^{\circ }-2x={180}^{\circ }\Rightarrow 2y-2x={180}^{\circ }-{360}^{\circ }\Rightarrow 2y-2x=-{180}^{\circ }\Rightarrow y-x=-{90}^{\circ }\Rightarrow x-y={90}^{\circ }$

Thus $\angle BAC-\angle OBC={90}^{\circ }$