The answer
(a) shown
(b) \(CD \approx 164\) m
(c) \(\approx \$46\,100\)
(d) \(\approx 15.4^{\circ}\)
O-Level E-Math 2015 Paper 2 Question 9 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 9 of the O-Level E-Math 2015 Paper 2. It tests pythagoras and tan in a right triangle, in the Trigonometry (2D & 3D) area. It is worth 13 marks: (a) 2, (b) 4, (c) 4, (d) 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.
In right-angled triangle \(ABD\) (right angle at \(A\)): \(BD = \sqrt{95^2 + 60^2} = \sqrt{12625} = 112.36\) m, and \(\angle ADB = \tan^{-1}\dfrac{95}{60} = 57.72^{\circ}\).
(a) Triangle \(BCD\) is isosceles (\(BC = CD\)) with apex \(\angle BCD = 40^{\circ}\), so \(\angle BDC = \dfrac{180^{\circ} - 40^{\circ}}{2} = 70^{\circ}\). Then \(\angle ADC = \angle ADB + \angle BDC = 57.72^{\circ} + 70^{\circ} = 127.7^{\circ}\) (1 d.p.). (shown)
(b) Sine rule in triangle \(BCD\): \(\dfrac{CD}{\sin \angle DBC} = \dfrac{BD}{\sin \angle BCD}\), so \(CD = \dfrac{112.36 \sin 70^{\circ}}{\sin 40^{\circ}} = 164\) m (3 s.f.).
(c) Area \(= \) area \(ABD + \) area \(BCD = \dfrac12(95)(60) + \dfrac12(CD)^2 \sin 40^{\circ} = 2850 + \dfrac12(164)^2 \sin 40^{\circ} = 2850 + 8672 = 11\,522 \text{ m}^2 = 1.1522\) hectares. Value \(= 1.1522 \times 40\,000 = \$46\,086 \approx \$46\,100\).
(d) Bird height above \(B\): \(\tan 18^{\circ} = \dfrac{h}{AB} \Rightarrow h = 95 \tan 18^{\circ} = 30.87\) m. The bird is vertically above \(B\), so the horizontal distance from \(D\) is \(DB = 112.36\) m. Angle of elevation from \(D = \tan^{-1}\dfrac{30.87}{112.36} = 15.4^{\circ}\) (1 d.p.).
Answer: (a) shown
(b) \(CD \approx 164\) m
(c) \(\approx \$46\,100\)
(d) \(\approx 15.4^{\circ}\)
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 pythagoras and tan in a right triangle question from Trigonometry (2D & 3D), worth 13 marks: (a) 2, (b) 4, (c) 4, (d) 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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