Question

In Fig. 8, 0 is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle.

Answer 1

Answer:

6.67 cm

Step-by-step explanation:

Given that PT is a tangent

⇒ ∠ OPT is a right angle

Find ∠ PTO:

sin (θ) = opp/hyp

sin (∠ PTO) = 5/13

∠ PTO = sin⁻¹ (5/13) = 22.62º

Find the length ET:

ET = OT - OE

ET = 13 - 5 = 8 cm

Given that AB is a tangent:

∠ AET is a right angle

Find the length AE:

tanθ = opp/adj

tan (22.62) = AE/8

AE = 10/3 cm

Find length AB:

AB = AE + EB

AB = 10/3 + 10/3 = 6.67 cm

Answer: Length AB = 6.67 cm

Answer 2

Here's the answer
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$\sf {\underline{Given :-}}$

OT = 13 cm
OP = OQ = 5 cm

Now,
PT = TQ

To find PT and TQ,

Using Pythagoras theorem,

PT = TQ = $\sf{ \sqrt{13 {}^{2} - 5 {}^{2} }}$

So,

PT = TQ = 12 cm

Now,

Let AB = x
BT = y

We know that ,
OT = 13 cm
OE = 5 cm

So,

TE = OT - OE = 8 cm

Now,

TE bisects AB

So,

AB = BE = $\sf{\frac{x}{2}}$

The length of tangents from a point of circle are equal.

So,

EB = QB = ${\sf{\frac{x}{2}}$

Now,

QT = QB + BT ... ( Q - B - T )

12 = x/2 + y

y = 12 - x/2... eq (1)

In ∆BET ,

$\sf{TE {}^{2} + BE {}^{2} = BT {}^{2} }$

We know that ,

ET = 8 cm
BE = x/2
BT = y

Also,
12 - x/2 = x

So,

$\sf{ {8}^{2} +( \frac{x}{2 } ){}^{2} = y {}^{2}}$

$\sf{64 + \frac{x {}^{2} }{4} = (12 - \frac{x}{2} = x ){}^{2}}$

$\sf \tiny{64 + \frac{x {}^{2} }{4} = 144 + \frac{x {}^{2} }{4} - 2 \times 12 \times \frac{x}{2}}$

$\sf{64 = 144 - 12x}$

$\sf{144 - 64 = 12x}$

$\sf{80 = 12x}$

$\sf{ \frac{80}{12} = x}$

Dividing L.H.S by 4,

$\sf{ \frac{20}{3} = x}$

Hence,

The length of AB = 20/3 cm or 6.67 cm

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Thanks!!