Question

Tp and tq are tangents to a circle with centre o at p and q respectively pq=8 cm and radius of a circle is 5cm find pq and tq

Answer 1

join OT now
PR=RQ=4cm
in triangle PQR
5²+4²=RO²
24-16=RO²
3=RO
angle TPR+angle RPO= 90= angle TPR+ angle PTR
angle RPO = angle PTR
TP/PO=RP/RO
TP = 3×5/3
TP= 20/3


marks as brainlest

Answer 2

Answer:

The value of TP and TQ is $\frac{20}{3}$ cm.

Step-by-step explanation:

Joint O to T.

Let it meet PQ at the point R.

Then ΔTPQ is an isosceles triangle and TO is the angle bisector of ∠PTO (Since TP = TQ = tangents from T on the circle)

Also, OT ⊥ PQ

Therefore, OT bisects PQ.

⇒ PR = RQ = 4 cm

Now,

$OR=\sqrt{OP^2-PR^2}$ (Using Pythagoras)

     $=\sqrt{5^2-4^2}$

     $=\sqrt{9}$

     $=3$ cm

Further,

∠TPR + ∠RPO = 90° (Since TPO = 90°)

Also, Since TRP = 90°

⇒ ∠TPR + ∠PTR = 90°

Thus, ∠RPO = ∠PTR

Therefore, right triangle TRP is similar to the right triangle PRO.

Hence $\frac{TP}{PO}=\frac{RP}{RO}$

$\frac{TP}{5}=\frac{4}{3}$

$TP=\frac{20}{3}$ cm.

Since TP = TQ.

So, $TQ=\frac{20}{3}$ cm.

Therefore, the value of TP and TQ is $\frac{20}{3}$ cm.

#SPJ3