Chords AC, AD, AB and BD are painted across a circular car park roundabout, all with endpoints on its edge. Angle ACB is \((x + 18)^{\circ}\) and angle ADB ...
Question 1 Report
Chords AC, AD, AB and BD are painted across a circular car park roundabout, all with endpoints on its edge. Angle ACB is \((x + 18)^{\circ}\) and angle ADB is \((2x - 6)^{\circ}\), both standing on the same arc AB.
- Angles in the same segment are equal; use this to construct an equation in \(x\), and solve it. (2)
- Determine the size of angle ADB. (1)
- Given that angle ABD is \((3x - 10)^{\circ}\), use the angle sum of triangle ABD to work out angle BAD. (2)
Answer Details
The circle theorem "angles in the same segment are equal" applies whenever two angles are subtended by the same chord from the same side of it, which gives an equation in \(x\) directly.
(a) Angles ACB and ADB both stand on chord AB from the same side, so they are equal: \( x+18=2x-6 \) [1 mark]. Subtracting \(x\) and adding \(6\) to both sides gives \( x=24 \) [1 mark].
(b) Substituting \( x=24 \): angle ADB \(=2(24)-6=42^\circ\) [1 mark].
(c) Given angle ABD \(=(3x-10)^\circ=3(24)-10=62^\circ\), the angle sum of triangle ABD gives angle BAD \(=180-\)angle \(ADB-\)angle \(ABD=180-42-62=76^\circ\) [2 marks].
As a check, angle ACB should also equal \(42^\circ\) (matching angle ADB, since both stand on the same arc AB): substituting \(x=24\) into \(x+18\) gives \(42\), confirming the value of \(x\) is consistent throughout.
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