- A65°
- B25°
- C40°
- D50°
Step-by-Step Solution
∠AOB = 180° − 50° = 130°. Triangle OAB is isosceles (OA = OB), so ∠OAB = (180° − 130°)/2 = 25°. (Equivalently ∠PAB = 65° and ∠OAB = 90° − 65°.)
Answer: B) 25°
PA and PB are tangents to a circle with centre O and ∠APB = 50°. Find ∠OAB.
Class 10Tangent Lengths and IncirclesGeometry
Master this chapter in 30 minutes a day.
Don't just memorise the answer — build a daily practice habit.
Practice 5 Similar Questions Free →All Tangent Lengths and Incircles practice questions
More Class 10 Tangent Lengths and Incircles practice
A triangle with perimeter 24 cm has an inscribed circle of radius 3 cm. What is the area…A quadrilateral ABCD circumscribes a circle. The tangent lengths from A, B, C and D are…A point P is at a distance from the centre equal to the diameter of the circle. Find the…Two straight roads from a village P touch a circular lake at A and B, and PA = 1.2 km. A…A circle is inscribed in a rhombus whose diagonals are 12 cm and 16 cm. Find the radius…