Circle Geometry, SMB-011

In the diagram, \(PR\) is a diameter of the circle centred \(O\) and \(\angle QPR = 15^{\circ} \).

Find \(\theta\). (2 marks)

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\(\theta = 75^{\circ}\)

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\(\angle PQR = 90^{\circ}\ \ \text{(angle in semicircle)} \)

\(\theta\)	\(=180-(90+15)\ \ (180^{\circ}\ \text{in}\ \Delta) \)
	\(=75^{\circ} \)