Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Step-by-Step Solution
Key Concept: In quadrilateral $OAPB$, $\angle OAP = 90^\circ, \angle OBP = 90^\circ$. Sum of angles $= 360^\circ \Rightarrow \angle APB + \angle AOB + 180^\circ = 360^\circ \Rightarrow \angle APB + \angle AOB = 180^\circ$.
In quadrilateral $OAPB$, $\angle OAP = \angle OBP = 90^\circ$. [1.0 Mark]
$\angle APB + \angle AOB + 90^\circ + 90^\circ = 360^\circ \Rightarrow \angle APB + \angle AOB = 180^\circ$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Identifying right angles $\angle OAP = \angle OBP = 90^\circ$: 1.0 Mark
Summing quadrilateral angles to get $180^\circ$: 1.0 Mark
Correct Answer: