Question

If tangent PA and PB from a point P to a circle with
centre o are inclined to each other at an angle of
80°, then POA is equal to :​

Answer 1

Step-by-step explanation:

• The diagram of the given question has been drawn in the attachment.

• From the diagram

=> Δ PAO  is congruent to Δ PBO.

=> as PO is common in Δ

=> ∠ PAO = ∠PBO = 90°

=> PA =PB [ tangent from an external point]

=> ∠POA= ∠POB

=> Also, ∠APB+ ∠AOB = 180°

=> 80° + ∠AOB = 180°

=> ∠AOB = 100°

=> ∠POA= ∠POB = 50°

Answer 2

Answer:

OA , OB are radii ,

PA , PB are tangents

∴∠PAO=90

∘

,∠PBO=90

∘

,∠BPA=80

o

In quadrilateral OAPB,

90

o

+90

o

+80

o

+∠AOB=360

=360−180

∘

−80

∘

=100

∘

But ∠POA=

2

1

∠AOB

=

2

1

×100 °

=50 °

∴∠POA=50 °