Question

AB is tangent to the circle with centre O at A and $OA=AB$
Find $\angle ABO$.

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

Answer: (B)

${45}^{\circ }$

In $△OAB$,

$OA=AB$ ....given

$\therefore △OAB$ is isosceles.

$\therefore \angle OBA=\angle AOB=x$ ....($\because OA=AB)$

$\angle OAB+\angle ABO+\angle AOB={180}^o$ ...Angle sum property o triangle

$\therefore {90}^{\circ }+x+x={180}^{\circ }$

$\Rightarrow x={45}^{\circ }$

So, option B is correct.