Question

solve this problem please

Answer 1

Given a circle with centre, O and two external tangents TP and TQ.

To prove that <PTQ = 2<OPQ.

Let PQ and OT intersect at R.

Proof: In triangles OPT and OQT,

<OPT = <OQT = 90 deg.

OPTQ is a quadrilateral. Since the opposite angles <OPT = <OQT = 90 deg,

<PTQ + <POQ = 360–90–90 = 180 deg.

So OPTQ if a cyclic quadrilateral and OT is the diameter. PQ is a chord.

Triangles OPR and OPT are right angled triangles.

<PRO = <OPT = right angles

<POR = <POT, being common angle.

Hence the third <OPR = <PTO …(1)

Similarly, <OQR = <QTO …(2)

Add (1) and (2) to get <OPR+<OQR = <PTO+<QTO = <PTQ.

But <OPR=<OQR =<OPQ [OPQ is an isosceles triangle]

Therefore, 2<OPQ = <PTQ.