Question

Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$. Find $\angle PTQ-2\angle OPQ$.

Answer 1

The length of tangents drawn from an internal point to a circle are equal.

Thus, $TP=TQ$

$\therefore \angle TQP=\angle TPQ($Angle opposite to equal sides are equal)

As we know,
$PT$ is tangent, $\&$ $OP$ is radius.

$\therefore OP\perp TP($Tangent at any point of circle is perpendicular to the radius through point of contact.)

So, $\angle OPT=90°$

$\angle OPQ+\angle TPQ=90$

$\angle TPQ=90-\angle OPQ$

$In△PTQ$

$\angle TPQ+\angle TQP+\angle PTQ=180°$

$\angle TPQ+\angle TPQ+\angle PTQ=180°$

$\angle TPQ+\angle PTQ=180°($from $\left(1\right)\angle YQP=\angle TPQ)$

$2(90°-\angle OPQ)+\angle PTQ=180°$

$180°-2\angle OPQ+\angle PTQ=180°$

$\angle PTQ=180-180+2\angle OPQ$

$\angle PTQ=2\angle OPQ$