Question

If pA and PB are tangents from an external paint
P to a circle with centre o such that LAPB = 50°
then LOAB is equal to​

Answer 1

Step-by-step explanation:

Given PA & PB are tangent to the circle with center O.

PA=PB [length of tangent from external point to circle are equal]

In ΔPAB

PA=PB

∠PBA=∠PAB [isosceles triangle]

now ∠PAB+∠PBA+∠APB=180

o

[Angle sum prop]

2∠PAB=180−50=130

∠PBA=∠PAB=65

o

………..(1)

Now PA is tangent & OA is radius at point A.

∠OAP=90

o

[tangent at any point is ⊥ to radius]

∠OAB=∠OAP−∠PAB=90−65=25

o

Hence angle OAB is 25

o

.