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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)}`
`/_BAC + /_BQP = 180°\ \ (APQB\ text{is a cyclic quad})`
`/_CQP + /_BQP = 180°\ text{(}/_ CQB\ text{is a straight line)}`
`:.\ /_BAC = /_CQP`
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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\(\angle BCD= 98^{\circ}\)
\( \angle BCD + \angle DAB = 180^{\circ} \ \ \text{(opposite angles of cyclic quad)} \)
\(\angle BCD\) | \(=180-(62+20)\) | |
\(=98^{\circ} \) |