ABD is formed by a tangent and a secant intersecting outside of a circle. If minor arc AC = 72° and minor arc CD = 132°, what is the measure of ∠ABD? A) 30° B) 36° C) 42° D) 48°

Answer:(B) 36°Step-by-step explanation:AB is a secant to the circle and B D is the tangent.Consider O as the center of the circle.then ∠A O C =72° and ∠ COD =132°As OD ⊥ B D [ Line from the center to the point of contact of tangent are perpendicular.]∠ OD B= 90°In Δ A O CO A = O C [radii of same circle]∠ O A C = ∠ O C A [ If sides are equal angle opposite to them are equal]∠ O A C  + ∠ O C A +∠ A O C =180° [ sum of angles of triangle is 180°]2∠O AC+72° = 180°∠O AC = 108°÷2∠O AC = 54°In Δ A B D∠A B D + ∠A DB+ ∠BAD =180° [angle sum property of triangle]∠A B D +90° +54° =180°∠A B D = 180 - 144°∠A B D =36°Option (B) is correct.