In the diagram, BD is a diameter of the circle centre O, ∡ATB = 30° and ∡ABD = 48°.
Calculate,
(i) ∡ACD
(ii) ∡ACB
(iii) ∡BDC
(iv) ∡BKC
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Answer:
(i) ∡ACD = ∡ABD (angles subtended by the same arc)
∡ACD = 48°
(ii) ∡BCD = 90° (angle at semi-circle)
∡ACB = ∡BCD - ∡ACD
∡ACB = 90° - 48°
∡ACB = 42°
(iii) ∡BDC = ∡DBA - ∡BTD (exterior angle = sum of 2 interior angles)
∡BDC = 48° - 30°
∡BDC = 18°
(iv) ∡CBD = 90° - 18° = 72°
∡BKC = 180° - ∡BCK - ∡CBK
∡BKC = 180° - ∡ACB - ∡CBD
∡BKC = 180° - 42° - 72°
∡BKC = 66°
