A circle with centre \(O\) has two tangents \(TP\) and \(TQ\) drawn from an external point \(T\). Using the properties of tangents, demonstrate that \(\angle PTQ = 2\angle OPQ\).
Proof:
We know that, the length of tangents drawn from to a circle are .
Therefore, \(TP = \)
So, \(\Delta TPQ\) is an triangle.
Therefore, \(\angle TPQ=\angle\)
\(TPQ = \) \(^\circ - \frac{1}{2}\theta\)
Also, we know that, the tangent at any point of a circle is to the through the point of contact.
\(\angle OPT=\)
\(\angle OPQ =\angle OPT-\)
\(\angle PTQ =2\)