The answer
(a) reflex angle \(ABC = 237^{\circ}\)
(b) angle \(ACD = 97^{\circ} \neq 90^{\circ}\), so \(C\) is not on the semicircle
O-Level E-Math 2016 Paper 1 Question 15 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 15 of the O-Level E-Math 2016 Paper 1. It tests angles on a straight line, in the Angles & circle property area. It is worth 5 marks: 4 + 1. It is a worded / diagram-based question, so open your Ten-Year Series (TYS) or the official paper at this question, then follow our full worked solution below.
(a) Step 1: \(AEC\) is a straight line, so angles \(DEC\) and \(DEA\) are angles on a straight line: angle \(DEA = 180^{\circ} - 45^{\circ} = 135^{\circ}\) (angles on a straight line). Step 2: In triangle \(AED\), the angles sum to \(180^{\circ}\): angle \(EAD = 180^{\circ} - 135^{\circ} - 22^{\circ} = 23^{\circ}\) (angle sum of a triangle), using angle \(ADE = 22^{\circ}\). Step 3: \(BC \parallel AD\) with transversal \(AC\), so angle \(ACB = \) angle \(CAD = 23^{\circ}\) (alternate angles); here angle \(CAD = \) angle \(EAD = 23^{\circ}\) because \(E\) lies on \(AC\). Step 4: In triangle \(ABC\), angle \(ABC = 180^{\circ} - \) angle \(BAC - \) angle \(ACB = 180^{\circ} - 34^{\circ} - 23^{\circ} = 123^{\circ}\) (angle sum of a triangle). Step 5: reflex angle \(ABC = 360^{\circ} - 123^{\circ} = 237^{\circ}\) (angles at a point).
(b) Angle \(ACD\) is the angle that \(AD\) subtends at \(C\). In triangle \(CED\), angle \(DCE = 180^{\circ} - 45^{\circ} - 38^{\circ} = 97^{\circ}\) (angle sum), and since \(E\) lies on \(AC\), angle \(ACD = 97^{\circ}\). An angle in a semicircle is exactly \(90^{\circ}\), but \(AD\) subtends \(97^{\circ}\) at \(C\), not \(90^{\circ}\), so \(C\) cannot lie on the semicircle with diameter \(AD\) (in fact \(97^{\circ} > 90^{\circ}\) means \(C\) lies inside it).
Answer: (a) reflex angle \(ABC = 237^{\circ}\)
(b) angle \(ACD = 97^{\circ} \neq 90^{\circ}\), so \(C\) is not on the semicircle
Same structure, different numbers
Swap the constants, dress a quadratic as a length, hide a derivative inside an integral, and a student sees a brand new problem. The structure underneath is the same, and so is the method. Once a student can name the structure, a whole row of questions that look different start to open the same way.
That is where marks really leak: in choosing the method, not in the algebra that follows. We call it Lock and Key, name the lock, then the key follows.
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Genius Plus Academy · O-Level & IP Mathematics
Our O-Level E-Math tuition trains the same recognise-the-structure method these worked solutions show, taught by a team that has marked these papers for years. It runs within our weekly Secondary Math programme, Sec 1 to 4 and IP.
It is a angles on a straight line question from Angles & circle property, worth 5 marks: 4 + 1.
Yes. IP (Integrated Programme) schools teach the same O-Level Mathematics content; they just sequence it differently and set their own internal exams, so these worked solutions apply to IP students too.
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