Question

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to
(a) 50°
(b) 60°
(c) 70°
(d) 80°

Answer 1

$\bigstar \: \boxed{\sf{\color{purple}{Answer:}}}$

Given that,

PA and PB are two tangents a circle and ∠APB=80 0

To find that ∠POA=?

Construction:- join OA,OBandOP

Proof:- Since OA⊥PA and OB⊥PB

Then ∠OAP=90

0 and ∠OBP=90 0

In ΔOAP&ΔOBP

OA=OB(radius)

OP=OP(Common)

PA=PB(lengthsoftangentdrawnfromexternalpointisequal)

∴ ΔOAP≅ΔOBP(SSScongruency)

So, [∠OPA=∠OPB(byCPCT)]

So,

∠OPA= 21 ∠APB = 21×80 0 =40 0

In ΔOPA,

∠POA+∠OPA+∠OAP=180 0

∠POA+40

0+90 0 =180 0

∠POA+130 0 =180 0

∠POA=180 0

−130 0

∠POA=50 0

The value of ∠POA is 50 0

Hence, Proved!

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