Question

In the figure if PA and PB are tangents and AP=PB then angle APB is​

Answer 1

Step-by-step explanation:

Given : A circle with center O and PA and PB are tangents to circle from a common external point P to point A and B respectively and ∠APB = 50°

To find : ∠OAB

OA ⏊ AP and OB ⏊ PB [ As tangent to at any point on the circle is perpendicular to the radius through point of contact]

∠OBP = ∠OAP = 90° [1]

In Quadrilateral AOBP [ By angle sum property of quadrilateral]

∠OBP + ∠OAP + ∠AOB + ∠APB = 360°

90° + 90° + ∠AOB + 50° = 360°

∠AOB = 130°

[2] Now in △OAB

OA = OB [Radii of same circle]

∠OBA = ∠ OAB

[3] Also,

By angle sum property of triangle

∠OBA + ∠OAB + ∠AOB = 180°

∠OAB + ∠OAB + 130 = 180

[using 2 and 3]

2∠OAB = 50°

∠OAB = 25°

Answer 2

Answer:

APB=50°

Step-by-step explanation:

PA and PB are tangents to the circle with centre O such that ∠APB = 50°.