If $PA$ and $PB$ are tangents to a circle with centre $O$ from an external point $P$ such that $\angle APB = 50^\circ$, find $\angle OAB$.
Step-by-Step Solution
Key Concept: $\angle AOB = 180^\circ - 50^\circ = 130^\circ$. In $\Delta OAB$, $OA = OB \Rightarrow \angle OAB = \dfrac{180^\circ - 130^\circ}{2} = 25^\circ$.
$\angle AOB = 180^\circ - 50^\circ = 130^\circ$. [1.0 Mark]
In isosceles $\Delta OAB$, $\angle OAB = \dfrac{180^\circ - 130^\circ}{2} = 25^\circ$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Finding $\angle AOB = 130^\circ$: 1.0 Mark
Evaluating $\angle OAB = 25^\circ$: 1.0 Mark
Correct Answer: