Question

In the given figure, ABCD is a cyclic quadrilateral, OB is the radius, PB is the tangent at point B and $\angle OBC={30}^{\circ }$. AOC is a straight line passing through the centre O. Then, the value of x is:

• A. ${60}^{\circ }$
• B. ${30}^{\circ }$
• C. ${120}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is A

${60}^{\circ }$

In the given figrue,
$\angle OBP={90}^{\circ }$
and $\angle OBC={30}^{\circ }$

$\therefore \angle CBP={90}^{\circ }-{30}^{\circ }={60}^{\circ }\Rightarrow \angle CAB={60}^{\circ }$ $[\because \angle CAB=\angle CBP$, alternate segment of chord]
In $\Delta OAB,$
OA = OB
$\Rightarrow \angle OAB=\angle OBA={60}^{\circ }$

So, $\angle OAB+\angle OBA+\angle AOB={180}^{\circ }$ [Sum of angles of a triangle is ${180}^{\circ }$]
$\Rightarrow {60}^{\circ }+{60}^{\circ }+\angle AOB={180}^{\circ }\therefore \angle AOB={180}^{\circ }-{120}^{\circ }={60}^{\circ }$