Question : In the given figure, $\mathrm{O}$ is the centre of the circle and $\angle\mathrm{AOB}=130^{\circ}$. Find $\angle\mathrm{APB}$.
Option 1: $110^{\circ}$
Option 2: $115^{\circ}$
Option 3: $100^{\circ}$
Option 4: $95^{\circ}$
Correct Answer: $115^{\circ}$
Solution :
Given:
$\angle{AOB}=130^{\circ}$
To find $\angle{APB}$
The arc AB subtends $\angle{AOB}$ at the centre and $\angle{AQB}$ at a point Q of the remaining parts of a circle.
$\therefore$ $\angle{AOB} = 2\angle{AQB}$
⇒ $\angle{AQB} = \frac{130^\circ}{2} = 65^\circ$
In a cyclic quadrilateral,
$\angle{APB}+\angle{AQB} = 180^\circ$
$\therefore$ $\angle{APB} = 180^\circ - 65^\circ = 115^\circ$
Hence, the correct answer is $115^\circ$.
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