Radius of the circle is 3 cm and PQ = 4.8 cm.

PQ is a chord of the circle with centre O.

The tangents at P and Q intersect at point T

So, TP and TQ are tangents

OP and OT are joined.

Join OQ

We have two right triangles: △OPT and △OQT

Here,

OT = OT (Common)

PT = QT (tangents of the circle)

OP = OQ (radius of the same circle)

By Side – Side – Side Criterion

△OPT ≅△OQT

Therefore, ∠POT = ∠OQT

Again, from triangles △OPR and △OQR

OR = OR (Common)

OP = OQ (radius of the same circle)

∠POR = ∠OQR (from above result)

By Side – Angle – Side Criterion

△OPR ≅△OQT

Therefore, ∠ORP = ∠ORQ

Now,

∠ORP + ∠ORQ = 180°

(Sum of linear angles = 180 degrees)

∠ORP + ∠ORP = 180°

∠ORP = 90°

This implies, OR ⏊ PQ and RT ⏊ PQ

Also OR perpendicular from center to a chord bisects the chord,

PR = QR = PQ/2 = 4.8/2 = 2.4 cm

Applying Pythagoras Theorem on right triangle △OPR,

(OP)² = (OR)² + (PR)²

(3)² = (OR)² + (2.4)²

OR = 1.8 cm

Applying Pythagoras Theorem on right angled △TPR,

(PT)² = (PR)² + (TR)² …(1)

Also, OP ⏊ OT

Applying Pythagoras Theorem on right △OPT,

(PT)² + (OP)² = (OT)²

[(PR)² + (TR)² ] + (OP)² = (TR + OR)²

(Using equation (1) and from figure)

(2.4)² + (TR)² + (3)² = (TR + 1.8)²

4.76 + (TR)² + 9 = (TR)² + 2(1.8)TR + (1.8)²

13.76 = 3.6 TR + 3.24

TR = 2.9 cm [approx.]

From (1) => PT² = (2.4)² + (2.9)²

PT² = 4.76 + 8.41

or PT = 3.63 cm [approx.]

Answer: Length of PT is 3.63 cm.