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°
2 answers:
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 C
O 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.
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