O Level Elementary Mathematics Practice Paper 2

O-Level Elementary Mathematics Quiz - Geometry Trigonometry

ANSWER KEY AND MARKING SCHEME

Total Marks: 50

Section A: Short Answer (10 marks)

1. (a) $\tan \angle ACB = \dfrac{8}{6} = \dfrac{4}{3}$ ✓ [1 mark]

(b) $AC = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$ cm ✓ [1 mark]

2. Interior angle = $156^\circ$
Exterior angle = $180^\circ - 156^\circ = 24^\circ$ ✓
Number of sides = $\dfrac{360^\circ}{24^\circ} = 15$ ✓ [2 marks]

3. Angle at centre = $110^\circ$
Angle at circumference = $\dfrac{1}{2} \times 110^\circ = 55^\circ$ ✓✓ [2 marks]
Answer: $55^\circ$

4. Sum of angles = $180^\circ$
Ratio total = $2 + 3 + 5 = 10$ parts ✓
Largest angle = $\dfrac{5}{10} \times 180^\circ = 90^\circ$ ✓ [2 marks]
Answer: $90^\circ$

5. $\cos \theta = \dfrac{2}{5}$ ✓
$\theta = \cos^{-1}(0.4) = 66.4^\circ$ (to 1 d.p.) ✓ [2 marks]
Answer: $66.4^\circ$

Section B: Structured Questions (20 marks)

6. (a) Using cosine rule:
$PR^2 = 12^2 + 15^2 - 2(12)(15)\cos 72^\circ$ ✓
$PR^2 = 144 + 225 - 360 \times 0.3090$
$PR^2 = 369 - 111.24 = 257.76$ ✓
$PR = \sqrt{257.76} = 16.1$ cm (3 s.f.) ✓ [3 marks]

(b) Area = $\frac{1}{2} \times 12 \times 15 \times \sin 72^\circ$ ✓
$= 90 \times 0.9511 = 85.6$ cm$^2$ (3 s.f.) ✓ [2 marks]

7. Area of larger circle = $\pi(8^2) = 64\pi$ cm$^2$ ✓
Area of smaller circle = $\pi(5^2) = 25\pi$ cm$^2$
Area of shaded region = $64\pi - 25\pi = 39\pi$ cm$^2$ ✓
Probability = $\dfrac{39\pi}{64\pi} = \dfrac{39}{64}$ ✓ [3 marks]

8. (a) The radius $OA$ is perpendicular to the tangent $TA$ at the point of contact.
Therefore, angle $OAT = 90^\circ$. ✓ [1 mark]

(b) In quadrilateral $OATB$:
Angles $OAT = OBT = 90^\circ$ (tangent-radius property)
Angle $ATB = 52^\circ$ (given)
Sum of angles in quadrilateral = $360^\circ$ ✓
Angle $AOB = 360^\circ - 90^\circ - 90^\circ - 52^\circ = 128^\circ$ ✓ [2 marks]

(c) Angle at circumference = $\frac{1}{2} \times$ angle at centre
Angle $ACB = \frac{1}{2} \times 128^\circ = 64^\circ$ ✓✓ [2 marks]

9. (a) Diagram should show:

• Point $P$ with north line
• $PQ$ at bearing $055^\circ$, length 12 km labelled
• $QR$ at bearing $145^\circ$, length 9 km labelled
• Right angle or angle $PQR$ indicated ✓✓ [2 marks]

(b) Angle $PQR = 145^\circ - 55^\circ = 90^\circ$ ✓
Using Pythagoras: $PR^2 = 12^2 + 9^2 = 144 + 81 = 225$ ✓
$PR = 15$ km ✓ [3 marks]

(c) $\tan(\text{angle } QPR) = \dfrac{9}{12} = 0.75$ ✓
Angle $QPR = 36.9^\circ$
Bearing of $R$ from $P = 055^\circ + 36.9^\circ = 091.9^\circ$ ✓ [2 marks]
Answer: $091.9^\circ$

Section C: Problem Solving (20 marks)

10. (a) Opposite angles of a cyclic quadrilateral sum to $180^\circ$.
Angle $BCD = 180^\circ - 78^\circ = 102^\circ$ ✓✓ [2 marks]
Reason: Opposite angles of a cyclic quadrilateral are supplementary.

(b) Angle $ABC = 180^\circ - 95^\circ = 85^\circ$ ✓✓ [2 marks]

(c) Angle at centre $BOC = 130^\circ$
Angle at circumference $BAC = \frac{1}{2} \times 130^\circ = 65^\circ$ ✓✓ [2 marks]

11. (a) Diagram should show:

• Vertical tower $XY$ on horizontal ground
• Points $A$, $B$, $X$ collinear with $B$ between $A$ and $X$
• Distance $AB = 50$ m labelled
• Angle of elevation from $A = 32^\circ$
• Angle of elevation from $B = 48^\circ$ ✓✓ [2 marks]

(b) Let $h$ = height of tower, $d = BX$
From $A$: $\tan 32^\circ = \dfrac{h}{d + 50}$ → $h = (d + 50)\tan 32^\circ$ ✓
From $B$: $\tan 48^\circ = \dfrac{h}{d}$ → $h = d\tan 48^\circ$ ✓
Equating: $d\tan 48^\circ = (d + 50)\tan 32^\circ$ ✓
$d(1.1106) = (d + 50)(0.6249)$
$1.1106d = 0.6249d + 31.245$
$0.4857d = 31.245$
$d = 64.33$ m ✓
$h = 64.33 \times \tan 48^\circ = 64.33 \times 1.1106 = 71.4$ m (3 s.f.) ✓ [5 marks]

12. (a) Using cosine rule to check if angle $ABC > 90^\circ$:
$\cos \angle ABC = \dfrac{7^2 + 9^2 - 11^2}{2(7)(9)}$ ✓
$= \dfrac{49 + 81 - 121}{126} = \dfrac{9}{126} = \dfrac{1}{14}$ ✓
Since $\cos \angle ABC > 0$, angle $ABC < 90^\circ$...

Correction: For obtuse angle, $\cos$ must be negative.
$\cos \angle ABC = \dfrac{AB^2 + BC^2 - AC^2}{2(AB)(BC)} = \dfrac{7^2 + 9^2 - 11^2}{2(7)(9)} = \dfrac{49 + 81 - 121}{126} = \dfrac{9}{126} > 0$

Wait — this gives acute. Let me recalculate with correct side labelling:
$\cos B = \dfrac{a^2 + c^2 - b^2}{2ac}$ where $a = BC = 9$, $b = AC = 11$, $c = AB = 7$
$\cos B = \dfrac{9^2 + 7^2 - 11^2}{2(9)(7)} = \dfrac{81 + 49 - 121}{126} = \dfrac{9}{126} = \dfrac{1}{14}$ ✓

Since $\cos B = \frac{1}{14} > 0$, angle $B$ is acute, not obtuse.

Revised approach: Check angle $A$ or $C$:
$\cos A = \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{11^2 + 7^2 - 9^2}{2(11)(7)} = \dfrac{121 + 49 - 81}{154} = \dfrac{89}{154} > 0$ (acute)

$\cos C = \dfrac{a^2 + b^2 - c^2}{2ab} = \dfrac{9^2 + 11^2 - 7^2}{2(9)(11)} = \dfrac{81 + 121 - 49}{198} = \dfrac{153}{198} > 0$ (acute)

All angles are acute — this triangle is acute-angled. The question premise is flawed.

Corrected solution for marking purposes:
If the question intended an obtuse angle, side lengths should be adjusted. For the given sides, all angles are acute. Award marks for correct cosine rule application showing $\cos B = \frac{1}{14} > 0$, concluding angle $B$ is acute. ✓✓✓ [3 marks — accept correct reasoning]

(b) $\angle ABC = \cos^{-1}\left(\frac{1}{14}\right) = 85.9^\circ$ (1 d.p.) ✓✓ [2 marks]

(c) Area of triangle = $\frac{1}{2} \times BC \times AD$
Also, Area = $\frac{1}{2} \times AB \times BC \times \sin B$ ✓
$= \frac{1}{2} \times 7 \times 9 \times \sin 85.9^\circ = 31.5 \times 0.9972 = 31.41$ cm$^2$
$\frac{1}{2} \times 9 \times AD = 31.41$
$AD = \dfrac{31.41 \times 2}{9} = 6.98$ cm (3 s.f.) ✓ [2 marks]

END OF ANSWER KEY