Question

In the given figure, $AB$ and $AC$ are tangents to the circle at $B$ and $C$, respectively, and $O$ is the centre of the circle, then $x$ equals

• A. ${65}^{\circ }$
• B. ${32.5}^{\circ }$
• C. ${57.5}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is C ${57.5}^{\circ }$

$OC\perp DA$ $\therefore$ $\angle OCD={90}^{\circ }$
In $△OCD,\angle ODC={180}^{\circ }-({90}^{\circ }+{65}^{\circ })$
$={180}^{\circ }-{155}^{\circ }={25}^{\circ }$
Also $OB\perp AB,$ $\therefore$ $\angle OBA={90}^{\circ }$
In $△DBA,\angle DAB={180}^{\circ }-(\angle BDA+\angle DBA)$
$={180}^{\circ }-({25}^{\circ }+{90}^{\circ })$
$={180}^{\circ }-{115}^{\circ }={65}^{\circ }$
$\therefore$ $\angle OAB=\frac{1}{2}\times \angle CAB$(or $\angle DAB$)$=\frac{1}{2}\times {65}^{\circ }={32.5}^{\circ }$
($\therefore$ Tangents are equally inclined to the line joining the external pt. and the centre of the circle)
In $△OBA,X={180}^{\circ }-({90}^{\circ }+{32.5}^{\circ })$
$={180}^{\circ }-{122.5}^{\circ }={57.5}^{\circ }$