In the figure, if PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠AOB is equal to ______

#### Options

25°

30°

40°

50°

#### Solution

In the figure, if PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠AOB is equal to **25°**.

**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 ΔOAB

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°