Question

In the given figure, if $PA$ and $PB$ are tangents to the circle with centre $O$ such that $\angle APB={54}^{\circ },$ then $\angle OAB$ equals

• A. ${16}^{\circ }$
• B. ${18}^{\circ }$
• C. ${27}^{\circ }$
• D. ${36}^{\circ }$

Answer 1

The correct option is C ${27}^{\circ }$

Given, $PA$ and $PB$ are the tangents from the point P.
$\angle APB={54}^{\circ }$
Now, In quadrilateral AOBP
$\angle OAP=\angle OBP={90}^{\circ }$ (Angle between tangent and radius)
Sum of angles = 360
$\angle OAP+\angle OBP+\angle OAB+\angle APB=360$
$90+90+54+\angle AOB=360$
$\angle AOB=126$
Now, In $△OAB$
$OA=OB$ (Radius of circle)
$\angle OAB=\angle OBA$ (Isosceles triangle property)
Sum of angles = 180
$\angle OAB+\angle OBA+\angle AOB=180$
$2\angle OAB+126=180$
$\angle OAB={27}^{\circ }$