Question
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Answer
Given, $\angle\text{ACB}=40^\circ$
We know that, a segment subtends an angle to the circle is half the angle subtends to the centre.
$\therefore\angle\text{AOB}=2\angle\text{ACB}$
$\Rightarrow\angle\text{ACB}=\frac{\angle\text{AOB}}{2}$
$\Rightarrow40^\circ=\frac{1}{2}\angle\text{AOB}$
$\Rightarrow\angle\text{AOB}=80^\circ\ \ \ ...(\text{i})$
In $\triangle\text{AOB},\ \text{AO}=\text{BO}$ [both are the radius of a circle]
$\Rightarrow\angle\text{OBA}=\angle\text{OAB}\ \ ...(\text{ii})$ [angle opposite to the equal sides are equal]
We know that, the sum of all three angles in a triangle AOB is 180°.
$\therefore\angle\text{AOB}+\angle\text{OBA}+\angle\text{OAB}=180^\circ$
$\Rightarrow80^\circ+\angle\text{OAB}+\angle\text{OAB}=180^\circ$ [from Eqs. (i) and (ii)]
$\Rightarrow2\angle\text{OAB}=180^\circ-80^\circ$
$\Rightarrow2\angle\text{OAB}=100^\circ$
$\therefore\angle\text{OAB}=\frac{100^\circ}{2}=50^\circ$
We know that, a segment subtends an angle to the circle is half the angle subtends to the centre.
$\therefore\angle\text{AOB}=2\angle\text{ACB}$
$\Rightarrow\angle\text{ACB}=\frac{\angle\text{AOB}}{2}$
$\Rightarrow40^\circ=\frac{1}{2}\angle\text{AOB}$
$\Rightarrow\angle\text{AOB}=80^\circ\ \ \ ...(\text{i})$
In $\triangle\text{AOB},\ \text{AO}=\text{BO}$ [both are the radius of a circle]
$\Rightarrow\angle\text{OBA}=\angle\text{OAB}\ \ ...(\text{ii})$ [angle opposite to the equal sides are equal]
We know that, the sum of all three angles in a triangle AOB is 180°.
$\therefore\angle\text{AOB}+\angle\text{OBA}+\angle\text{OAB}=180^\circ$
$\Rightarrow80^\circ+\angle\text{OAB}+\angle\text{OAB}=180^\circ$ [from Eqs. (i) and (ii)]
$\Rightarrow2\angle\text{OAB}=180^\circ-80^\circ$
$\Rightarrow2\angle\text{OAB}=100^\circ$
$\therefore\angle\text{OAB}=\frac{100^\circ}{2}=50^\circ$
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