A pair of straight lines from an external point F intersects a circle at A and B (FA < FB), and touches the circle at C. O is the centre of the circle. Given that ∠ACF = 50º and ∠AFC = 30º, find ∠AOB.
Show answer & explanation
Correct answer: C
Explanation
The correct answer is option 3 i.e. 100º.
∠OCF = 90° (Tangent makes an angle of 90 with the radius of the circle)
OC = OA = OB (Radius of the circle)
In the given figure, In △ACF
∠AFC = 30°, ∠ACF = 50°
∠AFC + ∠ACF + ∠CAF = 180°
⇒ 30° + 50° + ∠CAF = 180°
⇒ ∠CAF = 180° - 80° = 100°
∠OCA = 90° - ∠ACF = 90° - 50°° = 40°°
∠OCA = ∠OAC = 40°
∠OAB = 180° - (∠CAF + ∠OAC) = 180° - (100 + 40°) = 40°
∠OAB = ∠OBA = 40°
So, In △OAB
∠OAB + ∠OBA + ∠OAB = 180°
40° + 40° + ∠OAB = 180°
∠OAB = 100°