PQ IS A CHORD OF LENGTH …

Given, Chord PQ = 4.8 cm , radius of circle = 3 cm

Also, The tangents at P and Q intersect at a point T as shown in the figure.We have to find the length of TP.

Let TR = y and TP= x
We know that the perpendicular drawn from the center to the chord bisects It.Hence, PR = RQ ....(1)

Since, PQ = 4.8

or, PR + RQ = 4.8

or, PR + PR = 4.8 [ from (1)]

So, PR = 2.4

Now, in right triangle POR, Using Pythagoras theorem, we have
PO$^2$ = OR$^2$ + PR$^2$
$\Rightarrow 3 ^ { 2 } = O R ^ { 2 } + ( 2.4 ) ^ { 2 }$
$\Rightarrow O R ^ { 2 } = 3.24$
$\Rightarrow O R = 1.8$
Now, in right triangle TPR ,By Using Pythagoras theorem, we have
$T P ^ { 2 } = T R ^ { 2 } + P R ^ { 2 }$
$\Rightarrow x ^ { 2 } = y ^ { 2 } + ( 2.4 ) ^ { 2 }$
$\Rightarrow x ^ { 2 } = y ^ { 2 } + 5.76$ ...(2)
Again, In right triangle TPQ By Using Pythagoras theorem, we have,
$T O ^ { 2 } = T P ^ { 2 } + P O ^ { 2 }$
$\Rightarrow ( y + 1.8 ) ^ { 2 } = x ^ { 2 } + 3 ^ { 2 }$
$\Rightarrow y ^ { 2 } + 3.6 y + 3.24 = x^2 + 9$
$\Rightarrow y ^ { 2 } + 3.6 y = x^2 + 5.76$......(3)
Solving (2) and (3), we get 
x = 4cm and y = 3.2cm
$\therefore$ TP = 4cm.