Question

In the given figure, $TP$ and $TQ$ are tangents drawn to a circle with centre $O$. Prove that $\angle PTQ=2\angle OPQ$.

Answer 1

$OP\perp PT$

$\therefore$ $\angle OPT={90}^o$

Let $\angle OPQ=x^o$

$\angle QPT={90}^o-x^o$

Similarly $\angle PQT={90}^o-x^o$

In $△PQT$

$\angle PTQ+\angle QPT+\angle PQT={180}^o$

$\angle PTQ+{90}^o-x^o+{90}^o-x^o={180}^o$

$\therefore \angle PTQ=2x^o$

$\therefore PTQ=2\angle OPQ$