Question

In the diagram, $CB$ and $CD$ are tangents to the circle with centre $O.$ $AOC$ is a straight line and $\angle OCB={34}^{\circ }.$ $\angle ABO$ equals.

• A. ${56}^{\circ }$
• B. ${28}^{\circ }$
• C. ${34}^{\circ }$
• D. ${32}^{\circ }$

Answer 1

Answer: (B)

${28}^{\circ }$

In $△AOB$
$OA=OB$ (Radius of the circle)
Hence, $\angle OBA=\angle OAB=x$ (Isosceles triangle property)
Also, $\angle OBC={90}^{\circ }$ (Angle between the radius an the tangent)
Now, In $△AOC$
$\angle ACB+\angle BAC+\angle ABC=180$
$34+x+x+90=180$
$124+2x=180$
$x={28}^{\circ }$
Thus, $\angle ABO={28}^{\circ }$