Question

In Fig., if PA and PB are tangents to the circle with centre 0 such that ∠APB = 50°, then find ∠AOB .
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Answer 1

Answer:

In □PAOB,

∠APB+∠PBO+∠AOB+∠PAO=360

∘

....Angle sum property of quadrilateral

⇒∠AOB=360

∘

−(50

∘

+90

∘

+90

∘

)=130

Answer 2

Answer:

∠OAB = 25°

Step-by-step explanation:

∠APB = 50° (Given)

∠PAO = 90° (Angle between tangent and radius)

∠PAB = ∠PBA

∠APB + ∠PBA + ∠PAB = 180° (Angle sum property)

50° + 2∠PAB = 180°

2∠PAB = 180° - 50°

2∠PAB = 130°

∠PAB = 130°/2

∠PAB = 65°

∠OAB = ∠PAO - ∠PAB

= 90° - 50°

= 25°

Therefore, ∠OAB = 25°