Question

Question 16
In figure, $\angle ACB={40}^{\circ }$. Find $\angle OAB$.

Answer 1

Given, $\Delta ACB={40}^{\circ }$
We know that the angle subtended by the segment at any point on the circle is half the angle subtended by it to the centre.
$\therefore \angle AOB=2\angle ACB$
$\Rightarrow \angle ACB=\frac{\angle AOB}{2}$
$\Rightarrow {40}^{\circ }=\frac{1}{2}\angle AOB$
$\Rightarrow \angle AOB={80}^{\circ }$
AO = OB [both are the radius of a circle]
$\Rightarrow \angle OBA=\angle OAB$ [angles opposite to the equal sides are equal]...........(i)

We know that, the sum of all three angles in a triangle AOB is ${180}^{\circ }$
$\therefore \angle AOB+\angle OBA+\angle OAB={180}^{\circ }$
$\Rightarrow {80}^{\circ }+\angle OAB+\angle OAB={180}^{\circ }[FromEqs.(i\left)\right]$
$\Rightarrow 2\angle OAB={180}^{\circ }-{80}^{\circ }$
$\Rightarrow 2\angle OAB={100}^{\circ }$
$\therefore \angle OAB=\frac{{100}^{\circ }}{2}={50}^{\circ }$