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
Question 1 Report
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
Answer Details
In the diagram, \(\overline{AD}\) is a diameter of the circle with centre \(O\), \(B\) lies on the circle, and \(BC\) is a tangent to the circle at \(B\). It is given that \(\angle OAB = 34^\circ\).
(a) \(\angle OAB\)
\(OA\) and \(OB\) are radii, so triangle \(OAB\) is isosceles and \(\angle OBA = \angle OAB\):
\[ \angle OAB = 34^\circ \](b) \(\angle OCB\)
Since \(AD\) is a diameter, the angle in the semicircle is a right angle:
\[ \angle ABD = 90^\circ \]The angle subtended at the centre is twice the angle at the circumference on the same arc \(BD\), so
\[ \angle BOC = \angle BOD = 2 \times \angle OAB = 2 \times 34^\circ = 68^\circ \]A tangent is perpendicular to the radius at the point of contact, so at \(B\):
\[ \angle OBC = 90^\circ \]In triangle \(OBC\), the angles sum to \(180^\circ\):
\[ \angle BOC + \angle OBC + \angle OCB = 180^\circ \] \[ 68^\circ + 90^\circ + \angle OCB = 180^\circ \] \[ \angle OCB = 180^\circ - 158^\circ = 22^\circ \]Download The App On Google Playstore
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