Question

In the given figure, $AB$ is a chord of the circle with centre $O$ and $PQ$ is a tangent at point $B$ of the circle. If $\angle AOB={110}^{\circ },$ then $\angle ABQ$ is

• A. ${45}^{\circ }$
• B. ${70}^{\circ }$
• C. ${55}^{\circ }$
• D. ${35}^{\circ }$

Answer 1

The correct option is C ${55}^{\circ }$

Given, $\angle AOB={110}^{\circ }$
In $△AOB$
$OA=OB$ (Radius of the circle)
Thus, $\angle OAB=\angle OBA$ (Isosceles triangle property)
Sum of angles of the triangle = 180
$\angle AOB+\angle OAB+\angle OBA=180$
$110+2\angle OBA=180$
$\angle OBA={35}^{\circ }$
Since, PQ is a tangent touching the circle at B.
Thus, $\angle OBQ={90}^{\circ }$
Now, $\angle ABQ+\angle OBA=90$
$\angle ABQ+35=90$
$\angle ABQ={55}^{\circ }$