Secondary 4 Elementary Mathematics Preliminary Examination Paper 4

TuitionGoWhere Preliminary Practice Paper — Elementary Mathematics Secondary 4

Answer Key — Version 4 of 5

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Section A

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Question 1 [2]

Using the cosine rule:

AC² = AB² + BC² − 2(AB)(BC)cos(∠ABC)

AC² = 8² + 10² − 2(8)(10)cos 52°

AC² = 64 + 100 − 160 × 0.6157

AC² = 164 − 98.505

AC² = 65.495

AC = √65.495

AC = 8.09 cm (3 s.f.)

[2 marks: 1 mark for correct cosine rule setup, 1 mark for correct answer to 3 s.f.]

Common error: Using sine rule when two sides and included angle are given. Cosine rule is required.

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Question 2 [2]

Let θ be the angle the ladder makes with the ground.

cos θ = adjacent / hypotenuse = 2.5 / 6.5

cos θ = 0.3846

θ = cos⁻¹(0.3846)

θ = 67° (nearest degree)

[2 marks: 1 mark for correct trigonometric ratio, 1 mark for correct answer]

Common error: Using sin instead of cos. The distance from the wall is adjacent to the angle with the ground.

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Question 3 [2]

Bearing of P from Q = 225°

Bearing of Q from P = 225° − 180° = 045°

[2 marks: 1 mark for understanding back-bearing concept, 1 mark for correct answer]

Common error: Adding 180° instead of subtracting (or vice versa). If bearing > 180°, subtract 180°; if bearing < 180°, add 180°.

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Question 4 [2]

Check using Pythagoras' theorem:

XY² + YZ² = 12² + 9² = 144 + 81 = 225

XZ² = 15² = 225

Since XY² + YZ² = XZ², by the converse of Pythagoras' theorem, triangle XYZ is right-angled.

The right angle is at Y (between sides XY and YZ, opposite the hypotenuse XZ).

[2 marks: 1 mark for showing Pythagoras check, 1 mark for identifying right angle at Y]

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Question 5 [2]

sin θ = 0.62

Principal value: θ = sin⁻¹(0.62) = 38.3°

Since sin is positive in the 1st and 2nd quadrants:

θ₁ = 38° (nearest degree)

θ₂ = 180° − 38° = 142° (nearest degree)

θ = 38° and 142°

[2 marks: 1 mark for principal value, 1 mark for both correct answers]

Common error: Giving only the first-quadrant answer. For 0° ≤ θ ≤ 360°, two solutions exist when sin θ = positive value.

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Question 6 [2]

Let M be the foot of the perpendicular from O to chord AB.

Since the perpendicular from the centre bisects the chord: AM = 10/2 = 5 cm.

In right triangle OMA:

OM² + AM² = OA²

OM² + 5² = 7²

OM² = 49 − 25 = 24

OM = √24

OM = 4.90 cm (3 s.f.)

[2 marks: 1 mark for halving chord, 1 mark for correct answer]

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Question 7 [2]

Draw the diagram: P to Q is 40 km East, Q to R is 25 km North.

tan(θ) = opposite/adjacent = 25/40 = 0.625

θ = tan⁻¹(0.625) = 32.0°

Bearing of R from P = 090° − 32.0° = 058° (nearest degree)

[2 marks: 1 mark for correct diagram/trig setup, 1 mark for correct bearing]

Common error: Forgetting to convert the angle to a bearing measured clockwise from North.

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Question 8 [2]

Area = ½ × PQ × QR × sin(∠PQR)

Area = ½ × 7.2 × 9.8 × sin 68°

Area = ½ × 7.2 × 9.8 × 0.9272

Area = ½ × 65.42

Area = 32.7 cm² (3 s.f.)

[2 marks: 1 mark for correct formula, 1 mark for correct answer]

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Question 9 [2]

(a) From point A: tan 38° = h/x → h = x tan 38° [1]

From point B: tan 22° = h/(x + 30) → h = (x + 30) tan 22°

(b) Equating: x tan 38° = (x + 30) tan 22°

x(1.280) = (x + 30)(0.800) [using tan 38° ≈ 0.7813, tan 22° ≈ 0.4040]

Wait — let me recalculate carefully:

x tan 38° = (x + 30) tan 22°

x(0.7813) = (x + 30)(0.4040)

0.7813x = 0.4040x + 12.121

0.3773x = 12.121

x = 32.13 m

h = 32.13 × tan 38° = 32.13 × 0.7813 = 25.10

h = 25.1 m (3 s.f.) [1]

[Total: 2 marks — 1 for equations, 1 for correct height]

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Question 10 [2]

Area of trapezium = ½ × (sum of parallel sides) × perpendicular height

Area = ½ × (14 + 8) × 5

Area = ½ × 22 × 5

Area = 55

Area = 55 cm²

[2 marks: 1 mark for correct formula, 1 mark for correct answer]

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Section B

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Question 11 [4]

(a) Using the cosine rule:

cos(∠BAC) = (AB² + AC² − BC²) / (2 × AB × AC)

cos(∠BAC) = (13² + 15² − 14²) / (2 × 13 × 15)

cos(∠BAC) = (169 + 225 − 196) / 390

cos(∠BAC) = 198 / 390 = 0.5077

∠BAC = cos⁻¹(0.5077) = 59.49°

∠BAC = 59° (nearest degree) [2]

(b) Using Heron's formula:

s = (13 + 14 + 15) / 2 = 21

Area = √[s(s−a)(s−b)(s−c)]

Area = √[21 × 8 × 7 × 6]

Area = √[7056]

Area = 84.0 cm² (3 s.f.) [2]

Alternative: Area = ½ × 13 × 15 × sin 59° = ½ × 195 × 0.8572 = 83.6 ≈ 84.0 cm²

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Question 12 [4]

(a) Arc length = rθ (where θ is in radians)

Arc length = 12 × 2.4

Arc length = 28.8 cm [2]

(b) Area of sector = ½ r²θ

Area = ½ × 12² × 2.4

Area = ½ × 144 × 2.4

Area = 72 × 2.4

Area = 172.8 cm² [2]

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Question 13 [4]

(a) In triangle SPH:

∠PSH = 28° − 12° = 16° (the angle between SP and the horizontal at S, considering the path is inclined at 12°)

∠PHS = 180° − 40° = 140° (angle between PH and the horizontal at H... Let me reconsider.)

At point S: angle of elevation to H is 28°. The path SP is inclined at 12° to horizontal.

So ∠PSH = 28° − 12° = 16°.

At point P: angle of elevation to H is 40°. The path SP is inclined at 12°.

∠SPH = 40° + 12° = 52° (the angle at P in triangle SPH, between PS and PH).

Wait — let me think more carefully.

The angle between SP (going uphill at 12°) and the horizontal is 12°. From P, the angle of elevation of H is 40° (above horizontal). So the angle between SP (extended) and PH is 40° + 12° = 52°.

∠PHS = 180° − 16° − 52° = 112°.

Using sine rule in triangle SPH:

PH / sin(∠PSH) = SP / sin(∠PHS)

PH / sin 16° = 150 / sin 112°

PH = 150 × sin 16° / sin 112°

PH = 150 × 0.2756 / 0.9272

PH = 41.34 / 0.9272

PH = 44.6 m (3 s.f.) [2]

(b) In right triangle with PH as hypotenuse:

Vertical height = PH × sin(40°)

Height = 44.6 × sin 40° = 44.6 × 0.6428

Height = 28.7 m (3.s.f.) [2]

Note: The vertical height from P to H is PH × sin(40°) since 40° is the angle of elevation from P.

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Question 14 [4]

(a) In right triangle ADB:

AD² + BD² = AB²

AD² + 6² = 10²

AD² + 36 = 100

AD² = 64

AD = 8 cm [2]

(b) In right triangle ADC:

tan(∠ACB) = AD / DC = 8 / 5 = 1.6

∠ACB = tan⁻¹(1.6) = 58.0°

∠ACB = 58° (nearest degree) [2]

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Question 15 [4]

(a) Bearing from X to Y = 055°. Bearing from Y to Z = 145°.

The angle between the direction XY (055°) and the direction YZ (145°) is 145° − 55° = 90°.

Therefore ∠XYZ = 90° because the difference in bearings is 90°. [1]

(b) Since ∠XYZ = 90°, triangle XYZ is right-angled at Y.

XZ² = XY² + YZ² = 18² + 24² = 324 + 576 = 900

XZ = √900 = 30

XZ = 30.0 km (3 s.f.) [1]

(c) In triangle XYZ (right-angled at Y):

tan(θ) = YZ / XY = 24 / 18 = 1.333

θ = tan⁻¹(1.333) = 53.13°

The bearing of Z from X = 055° + 53.13° = 108.13°

Bearing of Z from X = 108° (nearest degree) [2]

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Question 16 [4]

(a) The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

∠TAB = 34° (tangent-chord angle)

∠ACB = 58° (angle in alternate segment... wait, let me reconsider.)

Actually, ∠TAB = 34° is the angle between tangent TA and chord AB. By the alternate segment theorem, this equals the angle in the alternate segment, which is ∠ACB.

But we're given ∠ACB = 58° and ∠TAB = 34°. These are not equal, so the angle in the alternate segment corresponding to ∠TAB would be the angle subtended by chord AB in the opposite segment.

Let me reconsider the geometry. ∠TAB is between tangent TA and chord AB. By alternate segment theorem, ∠TAB = angle in alternate segment subtended by chord AB = ∠ACB. But 34° ≠ 58°, so perhaps the diagram has C in a different position.

Given the problem as stated, let's work with what's provided:

Since TA is tangent at A, and OA is the radius to the point of contact:

OA ⊥ TA, so ∠OAT = 90°.

∠OAB = ∠OAT − ∠TAB = 90° − 34° = 56°

Reason: The radius to the point of contact is perpendicular to the tangent (∠OAT = 90°). [2]

(b) In triangle OAB, OA = OB (radii), so triangle OAB is isosceles.

∠OAB = ∠OBA = 56°

∠AOB = 180° − 56° − 56° = 68° [2]

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Section C

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Question 17 [5]

(a) In triangle DAB, using the cosine rule:

BD² = DA² + AB² − 2(DA)(AB)cos(∠DAB)

BD² = 70² + 120² − 2(70)(120)cos 74°

BD² = 4900 + 14400 − 16800 × 0.2756

BD² = 19300 − 4630.7

BD² = 14669.3

BD = √14669.3

BD = 121 m (nearest metre) [2]

(b) In triangle BCD, using the cosine rule:

cos(∠CBD) = (BC² + BD² − CD²) / (2 × BC × BD)

cos(∠CBD) = (85² + 121² − 95²) / (2 × 85 × 121)

cos(∠CBD) = (7225 + 14641 − 9025) / 20570

cos(∠CBD) = 12841 / 20570 = 0.6243

∠CBD = cos⁻¹(0.6243) = 51.36°

∠CBD = 51° (nearest degree) [2]

(c) Area of triangle DAB = ½ × DA × AB × sin(∠DAB)

= ½ × 70 × 120 × sin 74°

= ½ × 70 × 120 × 0.9613

= 4200 × 0.9613 = 4037.5 m²

Area of triangle BCD = ½ × BC × BD × sin(∠CBD)

= ½ × 85 × 121 × sin 51.36°

= ½ × 85 × 121 × 0.7807

= 5142.5 × 0.7807 = 4013.0 m²

Total area = 4037.5 + 4013.0 = 8050.5

Total area = 8051 m² (nearest m²) [1]

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Question 18 [5]

(a) Point A is due South of O. In right triangle OAT:

tan 48° = h / OA

h = OA × tan 48° [1]

(b) Point B is 80 m due West of A. So triangle OAB is right-angled at A, with OA (South) and AB = 80 m (West).

OB² = OA² + AB² = OA² + 80² = OA² + 6400

From point B, angle of elevation of T is 35°:

tan 35° = h / OB

h = OB × tan 35°

From (a): h = OA × tan 48°

So: OA × tan 48° = OB × tan 35°

OA × 1.1106 = √(OA² + 6400) × 0.7002

1.1106 × OA = 0.7002 × √(OA² + 6400)

Divide both sides by 0.7002:

1.5863 × OA = √(OA² + 6400)

Square both sides:

2.5163 × OA² = OA² + 6400

1.5163 × OA² = 6400

OA² = 4220.9

OA = √4220.9

OA = 65.0 m (3 s.f.) [2]

(c) h = OA × tan 48° = 65.0 × 1.1106

h = 72.2 m (3 s.f.) [2]

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Question 19 [6]

(a) Using the cosine rule in triangle PQR:

cos(∠PQR) = (PQ² + QR² − PR²) / (2 × PQ × QR)

cos(∠PQR) = (200² + 170² − 150²) / (2 × 200 × 170)

cos(∠PQR) = (40000 + 28900 − 22500) / 68000

cos(∠PQR) = 46400 / 68000 = 0.6824

∠PQR = cos⁻¹(0.6824) = 47.0°

∠PQR = 47° (nearest degree) [2]

(b) Area of triangle PQR using the sine formula:

Area = ½ × PQ × QR × sin(∠PQR)

Area = ½ × 200 × 170 × sin 47°

Area = 17000 × 0.7314 = 12433.1 m²

Also, Area = ½ × PR × QS

12433.1 = ½ × 150 × QS

12433.1 = 75 × QS

QS = 12433.1 / 75 = 165.77

QS = 166 m (nearest metre) [2]

(c) Area = 12433.1 m² (from part b)

Value = 12433.1 × $85 =$1,056,813.50

Total value = $1,056,814 (nearest dollar) [2]

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Mark Summary

Section	Marks
A (Q1–Q10)	20
B (Q11–Q16)	24
C (Q17–Q19)	16
Total	60