Question

For a circle with centre O, two tangents PA and PB form an angle of 80° from

an external point P. find m POA.​

Answer 1

Answer:

it should be 50°

180-40-90=50

Answer 2

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=

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.