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 \(180^\circ\):

\[ \angle BOC + \angle OBC + \angle OCB = 180^\circ \] \[ 68^\circ + 90^\circ + \angle OCB = 180^\circ \] \[ \angle OCB = 180^\circ - 158^\circ = 22^\circ \]