Question

Q.18. If Tangents PA and PB from point P to a circle with centre O are inclined to each other at an angle of

70°, then ZPOA is equal to

(a) 70°

(b) 55°

(c) 100°

(d) 40°​

Answer 1

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=

2

1

∠APB

=

2

1

×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

.