Question

11) Find the angle form at the centre of the circle if angle inclined by the tangents is 500.

Answer 1

Step-by-step explanation:

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

.