If tangents $PA$ and $PB$ from a point $P$ to a circle with center $O$ are inclined to each other at an angle of $80^\circ$, then $\angle POA$ is equal to:
(a) $50^\circ$
(b) $60^\circ$
(c) $70^\circ$
(d) $80^\circ$
Step-by-Step Solution
Key Concept: $OP$ bisects $\angle APB$, so $\angle APO = 40^\circ$. In right $\Delta OAP$, $\angle POA = 90^\circ - 40^\circ = 50^\circ$.
$\angle APO = \dfrac{80^\circ}{2} = 40^\circ$. [0.5 Mark]
In right $\Delta OAP$: $\angle POA = 90^\circ - 40^\circ = 50^\circ$. [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Finding $\angle APO = 40^\circ$: 0.5 Mark
Finding $\angle POA = 50^\circ$: 0.5 Mark
Correct Answer: $50^\circ$