Question

In the given figure, O is the centre of the circle. If $\angle AOC={140}^o$, AB=AC $\angle OAB$ is

• A. $\angle OAB={35}^o$
• B. $\angle OAB={45}^o$
• C. $\angle OAB={20}^o$
• D. $\angle OAB={30}^o$

Answer 1

The correct option is A $\angle OAB={35}^o$

Given that $\angle AOC={140}^{\circ }$

$\angle ABC=\frac{140}{2}={70}^{\circ }$

Because WKT angle made by a chord aa any point on the circle is half the angle made by it at the center

In $△AOC$

$AO=CO=radius$

$⟹\angle OAC=\angle OCA$

And

$2\angle OCA=180-\angle AOC$

$⟹\angle OCA=\frac{40}{2}=20=\angle OAC$

In $△ABC$

$AB=BC$

$⟹\angle BAC=\angle BCA$

And

$2\angle BAC=180-\angle ABC$

$⟹\angle BAC=55$

$\angle OAB=\angle CAB-\angle CAO=55–20={35}^{\circ }$