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°
