smc-4240-30-Cyclic quads

Circle Geometry, SMB-018

In the diagram, `AB` is a diameter of a circle with centre `O`. The point `C` is chosen such that `Delta ABC` is acute-angled. The circle intersects `AC` and `BC` at `P` and `Q` respectively.

Why is `/_BAC = /_CQP`? (2 marks)

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`text(Proof)\ text{(See Worked Solutions)}`

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`/_BAC + /_BQP = 180°\ \ (APQB\ text{is a cyclic quad})`

`/_CQP + /_BQP = 180°\ text{(}/_ CQB\ text{is a straight line)}`

`:.\ /_BAC = /_CQP`

Circle Geometry, SMB-014

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

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\(x= 84^{\circ}\)

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\(\angle ABC = 180-105=75^{\circ}\ \ \text{(straight line)} \)

\(\angle ADC\)	\(=180-75\ \ \text{(opposite angles of cyclic quad are supplementary)}\)
	\(=105^{\circ} \)

Circle Geometry, SMB-013

Find \(x\). (2 marks)

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\(x= 84^{\circ}\)

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\( \text{(Property: opposite angles of cyclic quad are supplementary)} \)

\(x+96\)	\(=180\)
\(x\)	\(=180-96\)
	\(=84^{\circ} \)

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

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