Answer:

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

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: ΔOPT and ΔOQT

Here,

OT=OT (Common)

PT=QT (tangents of the circle)

OP=OQ (radius of the same circle)

By Side - Side - Side Criterion

ΔOPT≅ΔOQT

Therefore, ∠POT=∠OQT

Again, from triangles ΔOPR and ΔOQR

OR=OR (Common)

OP=OQ (radius of the same circle)

∠POR=∠OQR (from above result)

By Side - Angle - Side Criterion

ΔOPR≅ΔOQT

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=2.4cm

Applying Pythagoras Theorem on right triangle ΔOPR,

(OP)

2

=(OR)

2

+(PR)

2

(3)

2

=(OR)

2

+(2.4)

2

OR=1.8cm

Applying Pythagoras Theorem on right angled ΔTPR,

(PT)

2

=(PR)

2

+(TR)

2

___(1)

Also, OP⊥OT

Applying Pythagoras Theorem on right ΔOPT,

(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=3.6TR+3.24

TR=2.9cm [approx.]

From (1) =PT

2

=(2.4)

2

+(2.9)

2

PT

2

=4.76+8.41

or PT=3.63cm [approx.]

Answer: Length of TP is 3.63

Answer:

TP = 4 c.m

Step-by-step explanation:

OP = 3

PR = 2.4

[tex]or = \sqrt{ {3}^{2} - {2.4}^{2} } [/tex]

OR = 1.8

take triangle OPR

[tex] \tan(o) = \frac{2.4}{1.8} [/tex]

[tex] \tan(o) = \frac{4}{3} [/tex]

take triangle POT

[tex] \tan(o) = \frac{tp}{po} [/tex]

[tex] \frac{4}{3} = \frac{tp}{3} [/tex]

So

TP = 4 c.m