Question

in the given figure AOB is the diameter of the circle with Centre O and AC is a tangent to the circle at A if angle BOC is equal to 130 then find angle is ACO

Answer 1

AOC + COB = 180 ( ANGLES IN A STRAIGHT LINE )
AOC = 180 - COB
AOC = 180 - 130
AOC = 50
AO = CO ( RADIUS OF A CIRCLE)
OAC = ACO = X ( BY LAW OF ISOSCELES TRIANGLE)
OAC + ACO + COA = 180 ( SUM OF ANGLE OF TRIANGLE)
X + X + 50 = 180
2X + 50 = 180
2X = 180 - 50
2X = 130
X = 65
THEREFORE , ACO = 65

Answer 2

Attached pic is created by ME

∠OAC = 90° (as radius ⊥ tangent)

∠BOC = ∠OAC + ∠ACO  (Exterior angle property)

⇒ 130° = 90° + ∠ACO

⇒ ∠ACO = 130° – 90°

= 40°

Thanks