Circle Geometry, SMB-012

In the diagram, \(AC\) is a diameter of the circle centred \(O\), \(\angle BAC = 20^{\circ}\) and \(\angle CAD = 62^{\circ} \).

Find \(\angle BCD\). (2 marks)

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\(\angle BCD= 98^{\circ}\)

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\( \angle BCD + \angle DAB = 180^{\circ} \ \ \text{(opposite angles of cyclic quad)} \)

\(\angle BCD\)	\(=180-(62+20)\)
	\(=98^{\circ} \)