Question

TP and TQ are tangents to a circle with cen tre O at P and Q respectively. PQ=8cm and radius of circle is 5 cm. Find TP and TQ.​

Answer 1

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Step-by-step explanation: Joint OT.  

Let it meet PQ at the point R.  

Then ΔTPQ is isosceles and TO is the angle bisector of ∠PTO.

[∵TP=TQ= Tangents from T upon the circle]

∴OT⊥PQ

∴OT bisects PQ.

PR=RQ=4 cm

Now,  

OR=  $\sqrt{op^{2} -pr^{2}$ =$\sqrt{5^{2} -4^{2}$   =3 cm

Now,  

∠TPR+∠RPO=90°   (∵TPO=90 °   )

=∠TPR+∠PTR(∵TRP=90   ° )

∴∠RPO=∠PTR

∴ Right triangle TRP is similar to the right triangle  

PRO. [By A-A Rule of similar triangles]

∴  $\frac{tp}{Po}$  = $\frac{rp}{ro}$  ⇒ $\frac{tp}{5}$  = $\frac{4}{3}$

⇒TP= $\frac{20}{3}$cm.