The answer
(a)(i) \(\angle EDC = 102^{\circ}\) (interior/co-interior angles, \(AE \parallel CD\))
(a)(ii) \(\angle BCD = 96^{\circ}\) (angle sum of a pentagon \(= 540^{\circ}\))
(b) both pairs of opposite sides of \(ACDE\) are parallel
O-Level E-Math 2020 Paper 1 Question 21 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 21 of the O-Level E-Math 2020 Paper 1. It tests co-interior angles in parallel lines, in the Angle properties / polygons area. It is worth 6 marks: (a)(i) 2, (a)(ii) 2, (b) 2. 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)(i) \(AE \parallel CD\) with transversal \(ED\), so \(\angle AED\) and \(\angle EDC\) are co-interior (allied) angles and add to \(180^{\circ}\): \(\angle EDC = 180^{\circ} - 78^{\circ} = 102^{\circ}\).
(ii) The interior angles of a pentagon sum to \((5-2)\times 180^{\circ} = 540^{\circ}\). Thus \(\angle BCD = 540^{\circ} - \angle EAB - \angle ABC - \angle AED - \angle EDC = 540^{\circ} - 120^{\circ} - 144^{\circ} - 78^{\circ} - 102^{\circ} = 96^{\circ}\).
(b) Triangle \(ABC\) is isosceles (\(AB = BC\)) with apex \(\angle ABC = 144^{\circ}\), so the base angles are \(\angle BAC = \angle BCA = \tfrac{180^{\circ} - 144^{\circ}}{2} = 18^{\circ}\). Then \(\angle EAC = \angle EAB - \angle BAC = 120^{\circ} - 18^{\circ} = 102^{\circ}\). Since \(\angle AED + \angle EAC = 78^{\circ} + 102^{\circ} = 180^{\circ}\), the converse of co-interior angles gives \(ED \parallel AC\). With \(AE \parallel CD\) already given, \(ACDE\) has both pairs of opposite sides parallel, so it is a parallelogram.
Answer: (a)(i) \(\angle EDC = 102^{\circ}\) (interior/co-interior angles, \(AE \parallel CD\))
(a)(ii) \(\angle BCD = 96^{\circ}\) (angle sum of a pentagon \(= 540^{\circ}\))
(b) both pairs of opposite sides of \(ACDE\) are parallel
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 co-interior angles in parallel lines question from Angle properties / polygons, worth 6 marks: (a)(i) 2, (a)(ii) 2, (b) 2.
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