The answer
(a) \(CX \approx 3.90\) cm
(b) angle \(XAB \approx 52.3^{\circ}\)
(c) area \(\approx 21.1\) cm²
O-Level E-Math 2019 Paper 2 Question 2 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 2 of the O-Level E-Math 2019 Paper 2. It tests cosine rule for a third side, in the Pythagoras & trigonometry area. It is worth 8 marks: 3 + 2 + 3. 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) In triangle \(BXC\) we know \(BX = 6.4\), \(BC = 8.3\) and the included angle \(\angle XBC = 27^{\circ}\). By the cosine rule, \[CX^2 = 6.4^2 + 8.3^2 - 2(6.4)(8.3)\cos 27° = 40.96 + 68.89 - 106.24\cos 27° = 109.85 - 94.66 = 15.19,\] so \(CX = \sqrt{15.19} = 3.897 \approx 3.90\) cm.
(b) In triangle \(AXB\) we know \(AB = 7.5\), \(BX = 6.4\) and the angle opposite \(AB\) is \(\angle AXB = 112^{\circ}\). By the sine rule, \[\frac{\sin \angle XAB}{BX} = \frac{\sin \angle AXB}{AB} \Rightarrow \sin \angle XAB = \frac{6.4 \sin 112°}{7.5} = \frac{6.4(0.9272)}{7.5} = 0.7912,\] so \(\angle XAB = \sin^{-1}(0.7912) = 52.30^{\circ} \approx 52.3^{\circ}\).
(c) Angle \(ABX = 180^{\circ} - 112^{\circ} - 52.30^{\circ} = 15.70^{\circ}\) (angle sum of triangle \(AXB\)). Then the full base angle \(\angle ABC = \angle ABX + \angle XBC = 15.70^{\circ} + 27^{\circ} = 42.70^{\circ}\). Using sides \(AB\) and \(BC\) with the included angle \(\angle ABC\), \[\text{area} = \tfrac12 (AB)(BC)\sin \angle ABC = \tfrac12 (7.5)(8.3)\sin 42.70° = 31.125 \times 0.6783 = 21.11 \approx 21.1 \text{ cm}^2.\]
Answer: (a) \(CX \approx 3.90\) cm
(b) angle \(XAB \approx 52.3^{\circ}\)
(c) area \(\approx 21.1\) cm²
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 cosine rule for a third side question from Pythagoras & trigonometry, worth 8 marks: 3 + 2 + 3.
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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