In the diagram, \(\overline{AD}\) is a diameter of a circle with Centre O. If ABD is a triangle in a semi-circle ∠OAB=34",
find: (a) ∠OAB (b) ∠OCB
∠OAB = ∠OBA=34°
∠OBA=90
∠OBD = ∠ABD - ∠0BA = 90 - 34 = 56
From triangle OBC
∠BOC + ∠OBC + ∠OCB= 180°
∠BOD =∠2OAB=68°
= 68° + 90° + ∠0CB=180
∠OCB= 180-158 =22
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}