Question

In above Figure, AB is a chord of a circle with centre O, AC is a tangent at A, making an angle of ${80}^{\circ }$ with AB. Then $\angle AOB$ is equal to:

• A. ${80}^{\circ }$
• B. ${50}^{\circ }$
• C. ${100}^{\circ }$
• D. ${160}^{\circ }$

Answer 1

Answer: (D)

${160}^{\circ }$

We know that, tangent to a circle is perpendicular to radius at the point of contact.

So, $\angle OAC=\angle OAB+\angle BAC$

$\Rightarrow \angle OAB+{80}^o={90}^o[\because \angle OAC={90}^o]$

$\Rightarrow \angle OAB={10}^o$

Also, $OA=OB$ [radii of same circle]

Hence, $△OAB$ is isosceles.

So, $\angle OAB=\angle OBA={10}^o$ [Angles opposite to equal sides of a $△$ are equal]

By using angle sum property of a triangle, we get:

$\angle AOB={180}^o-{10}^o-{10}^o$ $[\angle AOB+\angle OAB+\angle OBA={180}^o]$

$\Rightarrow \angle AOB={160}^o$

Hence, option D is correct