Question

In the figure, $\angle ACB={40}^o$. Find $\angle OAB$.

• A. ${100}^o$
• B. ${80}^o$
• C. ${60}^o$
• D. ${50}^o$

Answer 1

Answer: (D)

${50}^o$

In $\Delta OAB$, we have

$OA=OB$ .....(radii of the same circle).

$\therefore \Delta OAB$ is isosceles.
$\therefore \angle OAB=\angle OBA\Rightarrow \angle OAB+\angle OBA=2\angle OAB$

So, $\angle AOB+\angle OAB+\angle OBA=\angle AOB+2\angle OAB={180}^o$ ....(angle sum property of triangles).

$\therefore \angle OAB=\frac{1}{2}{(180}^o-\angle AOB)$ ..........(i)

Now $\angle AOB=2\angle ACB=2\times {40}^o={80}^o$ ....(The angle subtended by a chord of a circle at the centre is double of that at the cicumference)

From (i), we have

$\angle OAB=\frac{1}{2}{(180}^o-{80}^o)={50}^o$