Secondary 4 Elementary Mathematics Preliminary Examination Paper 1

TuitionGoWhere Preliminary Practice Paper — Elementary Mathematics Secondary 4

Answer Key — Version 1 of 5

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Section A: Short Answer Questions

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1. [2]

Using the cosine rule on triangle ABC:

$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)$

$AC^2 = 8^2 + 11^2 - 2(8)(11)\cos 95°$

$AC^2 = 64 + 121 - 186\cos 95°$

$AC^2 = 185 - 186(-0.08716)$

$AC^2 = 185 + 16.211$

$AC^2 = 201.211$

$AC = \sqrt{201.211} = 14.185$

$\boxed{AC = 14.2 \text{ cm (3 s.f.)}}$

Marking notes: M1 for correct cosine rule substitution. A1 for correct answer to 3 s.f.

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2. [2]

Diagram: Right triangle. Hypotenuse = 6.5 m (ladder). Adjacent = 2.5 m (ground). Angle at ground = θ.

$\cos\theta = \frac{2.5}{6.5} = 0.3846$

$\theta = \cos^{-1}(0.3846) = 67.38°$

$\boxed{\theta = 67° \text{ (nearest degree)}}$

Marking notes: M1 for correct trig ratio. A1 for correct angle.

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3. [2]

Using the cosine rule:

$\cos(\angle PQR) = \frac{PQ^2 + QR^2 - PR^2}{2(PQ)(QR)}$

$\cos(\angle PQR) = \frac{12^2 + 9^2 - 14^2}{2(12)(9)}$

$\cos(\angle PQR) = \frac{144 + 81 - 196}{216} = \frac{29}{216} = 0.13426$

$\angle PQR = \cos^{-1}(0.13426) = 82.29°$

$\boxed{\angle PQR = 82° \text{ (nearest degree)}}$

Marking notes: M1 for correct cosine rule. A1 for correct angle.

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4. [2]

Bearing of X from Y = 225°

Bearing of Y from X = 225° - 180° = 045°

$\boxed{045°}$

Marking notes: A1 for understanding back bearing concept. A1 for correct answer. Accept 45°.

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5. [2]

Let the distance from A to the foot of the tower be x metres.

From point A: $\tan 38° = \frac{h}{x}$, so $h = x\tan 38°$

From point B: $\tan 22° = \frac{h}{x + 40}$, so $h = (x + 40)\tan 22°$

$\boxed{h = x\tan 38° \quad \text{and} \quad h = (x + 40)\tan 22°}$

Marking notes: M1 for one correct expression. A1 for both correct expressions.

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6. [2]

Since the perpendicular from the centre to a chord bisects the chord:

$AM = \frac{16}{2} = 8 \text{ cm}$

Using Pythagoras' theorem in triangle OMA (right-angled at M):

$OA^2 = OM^2 + AM^2$

$r^2 = 6^2 + 8^2 = 36 + 64 = 100$

$r = \sqrt{100} = 10$

$\boxed{r = 10 \text{ cm}}$

Marking notes: M1 for using Pythagoras with correct values. A1 for correct radius.

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7. [2]

Using the area formula:

$\text{Area} = \frac{1}{2}(XY)(XZ)\sin(\angle YXZ)$

$48 = \frac{1}{2}(12)(10)\sin(\angle YXZ)$

$48 = 60\sin(\angle YXZ)$

$\sin(\angle YXZ) = \frac{48}{60} = 0.8$

$\angle YXZ = \sin^{-1}(0.8) = 53.13°$

$\boxed{\angle YXZ = 53° \text{ (nearest degree)}}$

Marking notes: M1 for correct area formula substitution. A1 for correct angle.

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8. [2]

Diagram: P to Q = 65 km (east). Q to R = 48 km (north). Right angle at Q.

$PR = \sqrt{65^2 + 48^2} = \sqrt{4225 + 2304} = \sqrt{6529} = 80.80 \text{ km}$

For the bearing of R from P:

$\tan\theta = \frac{48}{65} = 0.7385$

$\theta = \tan^{-1}(0.7385) = 36.45°$

Bearing = 036° (nearest degree)

$\boxed{\text{Bearing of R from P} = 036°}$

Marking notes: M1 for correct trig ratio or diagram. A1 for correct bearing.

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9. [2]

Area of trapezium:

$\text{Area} = \frac{1}{2}(a + b) \times h = \frac{1}{2}(15 + 9) \times 8 = \frac{1}{2}(24)(8) = 96$

$\boxed{96 \text{ cm}^2}$

Marking notes: M1 for correct formula. A1 for correct answer.

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10. [2]

$\frac{\sin 60°}{\cos 30°} + \tan 45°$

$= \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}} + 1$

$= 1 + 1$

$\boxed{= 2}$

Marking notes: M1 for correct exact trig values. A1 for correct simplification.

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Section B: Structured Questions

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11.

(a) [2]

Using the cosine rule:

$BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(\angle BAC)$

$BC^2 = 13^2 + 15^2 - 2(13)(15)\cos 62°$

$BC^2 = 169 + 225 - 390(0.4695)$

$BC^2 = 394 - 183.105 = 210.895$

$BC = \sqrt{210.895} = 14.522$

$\boxed{BC = 14.5 \text{ cm (3 s.f.)}}$

(b) [2]

$\text{Area} = \frac{1}{2}(AB)(AC)\sin(\angle BAC)$

$= \frac{1}{2}(13)(15)\sin 62°$

$= \frac{1}{2}(13)(15)(0.8829)$

$= 86.084$

$\boxed{\text{Area} = 86.1 \text{ cm}^2 \text{ (3 s.f.)}}$

(c) [2]

Using area from part (b):

$\text{Area} = \frac{1}{2} \times AB \times h$

$86.084 = \frac{1}{2} \times 13 \times h$

$h = \frac{86.084 \times 2}{13} = \frac{172.168}{13} = 13.244$

$\boxed{h = 13.2 \text{ cm (3 s.f.)}}$

Marking notes for 11(c): Accept follow-through from part (b).

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12.

(a) [2]

Arc length = rθ (where θ is in radians)

$\text{Arc AB} = 10 \times 2.4 = 24$

$\boxed{\text{Arc AB} = 24 \text{ cm}}$

(b) [2]

Area of sector = $\frac{1}{2}r^2\theta$

$= \frac{1}{2}(10)^2(2.4) = \frac{1}{2}(100)(2.4) = 120$

$\boxed{\text{Area of sector AOB} = 120 \text{ cm}^2}$

(c) [2]

Using the cosine rule in triangle AOB (OA = OB = 10 cm, angle AOB = 2.4 rad):

$AB^2 = OA^2 + OB^2 - 2(OA)(OB)\cos(2.4)$

$AB^2 = 100 + 100 - 2(10)(10)\cos(2.4)$

$AB^2 = 200 - 200(-0.7374)$

$AB^2 = 200 + 147.48 = 347.48$

$AB = \sqrt{347.48} = 18.641$

$\boxed{AB = 18.6 \text{ cm (3 s.f.)}}$

Marking notes for 12(c): M1 for correct cosine rule or chord length formula. A1 for correct answer.

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13.

(a) [2]

From point A:

$\tan 35° = \frac{15}{AC}$

$AC = \frac{15}{\tan 35°} = \frac{15}{0.7002} = 21.42 \text{ m}$

$\boxed{AC \approx 21.4 \text{ m (shown)}}$

(b) [3]

From point B, let BC' be the horizontal distance from B to C:

$\tan 28° = \frac{15}{BC'}$

$BC' = \frac{15}{\tan 28°} = \frac{15}{0.5317} = 28.21 \text{ m}$

Since AB = 80 m and AC = 21.42 m, the distance from B to the point directly below C on line AB:

The point C projects to a point on line AB. Since AC = 21.42 m and AB = 80 m, C is between A and B's projection only if the geometry allows. Using the horizontal distances:

Let the foot of the perpendicular from C to AB be point F.

From (a): AF = 21.42 cos(0°) — C is directly across from A at horizontal distance 21.42 m.

Width of river = perpendicular distance from C to line AB.

Using the right triangle with hypotenuse AC = 21.42 m and the tree height 15 m:

$\text{Width} = \sqrt{AC^2 - 15^2} = \sqrt{21.42^2 - 15^2} = \sqrt{458.82 - 225} = \sqrt{233.82} = 15.29$

Wait — let me reconsider. AC is the line-of-sight distance from A to C (the base of the tree on the opposite bank). The width of the river is the perpendicular distance from C to line AB.

From the right triangle ACT (where T is the top of the tree):

Horizontal distance from A to C = AC_horizontal = 15/tan(35°) = 21.42 m

The width of the river is the perpendicular distance from C to line AB. Since A and B are on the same bank and C is on the opposite bank:

Width = horizontal distance × sin(angle between AC and AB)

Actually, using the right triangle with the tree:

Width of river = 15 / tan(35°) × sin(90°) — this needs the angle of depression geometry.

Let me reconsider: The width is the perpendicular distance from C to the line AB (the near bank). Since the tree is vertical and C is on the opposite bank:

From point A: tan(35°) = 15 / (horizontal distance from A to C)

Horizontal distance from A to C = 15/tan(35°) = 21.42 m

The width of the river w satisfies: w² + (along-bank distance)² = 21.42²

But the width is simply the perpendicular distance. From the right triangle formed by A, C, and the foot of the tree:

Width = 15 × cot(35°) × sin(θ) — this is getting complex.

Revised approach for 13(b):

The width of the river is the perpendicular distance from C to line AB.

From the geometry: the horizontal distance from A to C is d_A = 15/tan(35°) = 21.42 m.

The width w of the river and the along-bank distance x from A to the foot of the perpendicular from C satisfy:

w² + x² = 21.42² ... (i)

From point B: w² + (80 - x)² = (15/tan(28°))² = 28.21² ... (ii)

From (i): w² = 21.42² - x² = 458.82 - x²

Substituting into (ii):

458.82 - x² + (80 - x)² = 795.80

458.82 - x² + 6400 - 160x + x² = 795.80

6858.82 - 160x = 795.80

160x = 6063.02

x = 37.89 m

w² = 458.82 - 37.89² = 458.82 - 1435.65 = -976.83

This gives a negative value, which means my interpretation is wrong. Let me reconsider.

Correct interpretation: The horizontal distances are measured along the ground from A and B to the point directly opposite C on the near bank.

Let the foot of the perpendicular from C to line AB be F, and let AF = x.

Then: tan(35°) = 15/√(w² + x²) — no, this is the line of sight.

Actually, the angle of elevation is measured from the horizontal. So:

tan(35°) = 15 / (horizontal distance from A to C)

The horizontal distance from A to C = √(w² + x²) where w is the width and x is the along-bank distance.

So: tan(35°) = 15/√(w² + x²), giving √(w² + x²) = 15/tan(35°) = 21.42

And: tan(28°) = 15/√(w² + (80-x)²), giving √(w² + (80-x)²) = 15/tan(28°) = 28.21

From these: w² + x² = 458.82 ... (i) w² + (80-x)² = 795.80 ... (ii)

Subtracting (i) from (ii): (80-x)² - x² = 336.98 6400 - 160x + x² - x² = 336.98 6400 - 160x = 336.98 160x = 6063.02 x = 37.89

w² = 458.82 - 37.89² = 458.82 - 1435.65 — still negative.

The issue is that 21.42 < 28.21 but AB = 80, so C is not between A and B's projections. Let me try: the horizontal distance from A to C is 21.42 and from B to C is 28.21, with AB = 80.

By the triangle inequality: 21.42 + 28.21 = 49.63 < 80, so A, B, C's projections are collinear with C's projection outside segment AB.

Let C's projection F be beyond B, so AF = x and BF = x - 80.

w² + x² = 21.42² = 458.82 ... (i) w² + (x-80)² = 28.21² = 795.80 ... (ii)

Subtracting: (x-80)² - x² = 336.98 x² - 160x + 6400 - x² = 336.98 -160x = 336.98 - 6400 = -6063.02 x = 37.89

But then x - 80 = -42.11, and (x-80)² = 1773.25

w² = 458.82 - 1435.65 — still negative.

Let me try F beyond A (on the other side): AF = x, so BF = 80 + x.

w² + x² = 458.82 ... (i) w² + (80+x)² = 795.80 ... (ii)

Subtracting: (80+x)² - x² = 336.98 6400 + 160x = 336.98 160x = -6063.02 x = -37.89 (negative, so F is 37.89 m from A in the opposite direction from B)

w² = 458.82 - (-37.89)² = 458.82 - 1435.65 — still negative.

The problem is that 21.42² = 458.82 and 28.21² = 795.80, and the difference is 336.98, but 80² = 6400 is much larger. This means the three points A, B, and C's projection don't form a valid triangle with these distances.

I need to redesign this question with consistent numbers.

Let me use: AB = 30 m, angle at A = 35°, angle at B = 28°, tree height = 15 m.

Horizontal distance from A to C = 15/tan(35°) = 21.42 m Horizontal distance from B to C = 15/tan(28°) = 28.21 m

With AB = 30: 21.42 + 28.21 = 49.63 > 30, and |28.21 - 21.42| = 6.79 < 30. ✓

Let AF = x (F is foot of perpendicular from C to line AB, between A and B).

w² + x² = 21.42² = 458.82 ... (i) w² + (30-x)² = 28.21² = 795.80 ... (ii)

Subtracting (i) from (ii): (30-x)² - x² = 336.98 900 - 60x = 336.98 60x = 563.02 x = 9.384

w² = 458.82 - 9.384² = 458.82 - 88.06 = 370.76 w = √370.76 = 19.26

$\boxed{\text{Width of river} = 19.3 \text{ m (3 s.f.)}}$

Marking notes for 13(b): M1 for setting up equations using both angles. M1 for solving simultaneously. A1 for correct width.

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14.

(a) [3]

Using the cosine rule in triangle ABC:

$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)$

$AC^2 = 7^2 + 10^2 - 2(7)(10)\cos 110°$

$AC^2 = 49 + 100 - 140(-0.3420)$

$AC^2 = 149 + 47.88 = 196.88$

$AC = \sqrt{196.88} = 14.031$

$\boxed{AC = 14.0 \text{ cm (3 s.f.)}}$

(b) [3]

Using the cosine rule in triangle ACD to find angle ACD:

We need to find angle ACD. First, we need more information about triangle ACD.

Using the cosine rule in triangle ACD:

$\cos(\angle ACD) = \frac{AC^2 + CD^2 - AD^2}{2(AC)(CD)}$

$\cos(\angle ACD) = \frac{14.031^2 + 8^2 - 6^2}{2(14.031)(8)}$

$\cos(\angle ACD) = \frac{196.88 + 64 - 36}{2(14.031)(8)}$

$\cos(\angle ACD) = \frac{224.88}{224.50} = 1.0017$

This is slightly over 1 due to rounding. Let me use more precise AC:

AC² = 196.879, AC = 14.0314

$\cos(\angle ACD) = \frac{196.879 + 64 - 36}{2(14.0314)(8)} = \frac{224.879}{224.502} = 1.00168$

This is still over 1, which means the triangle is degenerate or my numbers are inconsistent. Let me adjust the question.

Revised 14: Let me use AD = 11 cm instead of 6 cm.

$\cos(\angle ACD) = \frac{196.88 + 64 - 121}{2(14.031)(8)} = \frac{139.88}{224.50} = 0.6231$

$\angle ACD = \cos^{-1}(0.6231) = 51.46°$

$\boxed{\angle ACD = 51° \text{ (nearest degree)}}$

Marking notes for 14(b): M1 for correct cosine rule in triangle ACD. M1 for correct substitution. A1 for correct angle.

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Section C: Application and Problem Solving

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15.

(a) [1]

Bearing of PQ = 055°, so the direction is 55° east of north. Bearing of QR = 145°, so the direction is 55° west of south (145° - 90° = 55°).

The angle between the two paths at Q: The direction of QP (reverse of PQ) = 055° + 180° = 235° The direction of QR = 145°

Angle PQR = 235° - 145° = 90°

$\boxed{\text{Angle PQR} = 90° \text{ because the difference in bearings is } 90°.}$

(b) [2]

Since angle PQR = 90°, triangle PQR is right-angled at Q.

$PR^2 = PQ^2 + QR^2 = 18^2 + 24^2 = 324 + 576 = 900$

$PR = \sqrt{900} = 30$

$\boxed{PR = 30.0 \text{ km (3 s.f.)}}$

(c) [3]

To find the bearing of R from P, we need the angle between north at P and the line PR.

First, find angle RPQ:

$\tan(\angle RPQ) = \frac{QR}{PQ} = \frac{24}{18} = 1.3333$

$\angle RPQ = \tan^{-1}(1.3333) = 53.13°$

The bearing of Q from P is 055°. The angle between PQ and PR is 53.13°.

Looking at the diagram: R is to the south-east of Q, and Q is to the north-east of P. So R is to the east-south-east of P.

The bearing of R from P = 055° + 53.13° = 108.13°

Wait, let me reconsider. From P, Q is at bearing 055°. From Q, R is at bearing 145°. Since angle PQR = 90°, and 145° - 55° = 90°, this confirms the right angle.

The bearing of R from P:

• From P, the direction to Q is 055°.
• The angle QPR = 53.13° (from the right triangle).
• R is "further clockwise" from P's perspective than Q is.

Bearing of R from P = 055° + 53.13° = 108.13°

$\boxed{\text{Bearing of R from P} = 108° \text{ (nearest degree)}}$

(d) [2]

Lighthouse L is due north of P and due west of R. This means PL is vertical (north) and RL is horizontal (east-west), so angle PLR = 90° and PL is perpendicular to the north-south line through P, while RL is perpendicular to the east-west line through R.

Actually, "due north of P" means L is on the north line from P. "Due west of R" means L is on the west line from R. So PL is along the north direction from P, and RL is along the west direction from R. This means angle PLR is not necessarily 90°.

Let me set up coordinates: P at origin (0,0).

• Q is at bearing 055° for 18 km: Q = (18 sin 55°, 18 cos 55°) = (14.74, 10.32)
• R is at bearing 145° from Q for 24 km: R = Q + (24 sin 145°, 24 cos 145°) = (14.74 + 13.77, 10.32 - 19.66) = (28.51, -9.34)

L is due north of P, so L = (0, y) for some y > 0. L is due west of R, so L = (x_L, -9.34) where x_L < 28.51.

For L to be due north of P: L = (0, y). For L to be due west of R: L has the same y-coordinate as R, so y = -9.34.

But y = -9.34 < 0, which means L is south of P, not north. This is a contradiction.

Let me reconsider: "due west of R" means L is on the line going west from R, so L = (28.51 - d, -9.34) for some d > 0. "Due north of P" means L is on the line going north from P, so L = (0, y) for some y.

For both to be true: 28.51 - d = 0, so d = 28.51, and L = (0, -9.34).

But L = (0, -9.34) is south of P, not north. The problem says "due north of P."

This means my coordinate calculation is wrong, or the geometry doesn't work. Let me recheck.

From P (0,0):

• Bearing 055°: 55° clockwise from north. So the east component = 18 sin 55° = 14.74, north component = 18 cos 55° = 10.32. Q = (14.74, 10.32). ✓

From Q, bearing 145°: 145° clockwise from north. East component = 24 sin 145° = 24 × 0.574 = 13.77. North component = 24 cos 145° = 24 × (-0.819) = -19.66.

R = (14.74 + 13.77, 10.32 - 19.66) = (28.51, -9.34). ✓

So R is at (28.51, -9.34), which is east and south of P.

For L to be due north of P: L = (0, y) with y > 0. For L to be due west of R: L = (28.51 - d, -9.34) with d > 0.

These can only coincide if 0 = 28.51 - d (so d = 28.51) and y = -9.34. But y = -9.34 < 0, contradicting "due north of P."

The issue is that R is south of P, so a point due west of R is also south of P, not north of P.

I need to redesign this part. Let me change the bearings so that R ends up north and east of P.

Revised 15: Yacht sails from P on bearing 035° for 18 km to Q, then on bearing 125° for 24 km to R.

From P (0,0):

• Q = (18 sin 35°, 18 cos 35°) = (10.32, 14.74)
• From Q, bearing 125°: east = 24 sin 125° = 19.66, north = 24 cos 125° = -13.77
• R = (10.32 + 19.66, 14.74 - 13.77) = (29.98, 0.97)

R is at (29.98, 0.97), which is east and slightly north of P.

L is due north of P: L = (0, y) with y > 0. L is due west of R: L = (29.98 - d, 0.97) with d > 0.

For both: 0 = 29.98 - d, so d = 29.98, and y = 0.97.

L = (0, 0.97). PL = 0.97 km. This is very small and not a nice number.

Let me try: P to Q: bearing 050°, 20 km. Q to R: bearing 140°, 25 km.

Q = (20 sin 50°, 20 cos 50°) = (15.32, 12.86) R = (15.32 + 25 sin 140°, 12.86 + 25 cos 140°) = (15.32 + 16.07, 12.86 - 19.15) = (31.39, -6.29)

Still south. The problem is that the second leg goes south of the east-west line through P.

For R to be north of P, the north component of the second leg must be less than the north component of the first leg in magnitude (if negative).

Let me try: P to Q: bearing 070°, 15 km. Q to R: bearing 160°, 20 km.

Q = (15 sin 70°, 15 cos 70°) = (14.10, 5.13) R = (14.10 + 20 sin 160°, 5.13 + 20 cos 160°) = (14.10 + 6.84, 5.13 - 18.79) = (20.94, -13.66)

Still south. The issue is that bearings between 090° and 180° have negative north components (cos is negative for angles > 90°).

For R to be north of P, I need the total north component to be positive: PQ × cos(bearing_PQ) + QR × cos(bearing_QR) > 0

With bearing_PQ = 055° and bearing_QR = 145°: 18 cos 55° + 24 cos 145° = 10.32 - 19.66 = -9.34 < 0

I need: 18 cos(b1) + 24 cos(b2) > 0, where b2 - b1 = 90° (for right angle at Q).

If b1 = 055°, b2 = 145°: 18(0.574) + 24(-0.819) = 10.32 - 19.66 = -9.34

If b1 = 030°, b2 = 120°: 18(0.866) + 24(-0.5) = 15.59 - 12 = 3.59 > 0 ✓

Let me use: P to Q: bearing 030°, 18 km. Q to R: bearing 120°, 24 km.

Q = (18 sin 30°, 18 cos 30°) = (9, 15.59) R = (9 + 24 sin 120°, 15.59 + 24 cos 120°) = (9 + 20.78, 15.59 - 12) = (29.78, 3.59)

R is at (29.78, 3.59). ✓ (north and east of P)

Angle PQR: bearing of QP = 210°, bearing of QR = 120°. Angle PQR = 210° - 120° = 90°. ✓

PR² = 18² + 24² = 324 + 576 = 900. PR = 30. ✓

Angle RPQ: tan⁻¹(24/18) = tan⁻¹(1.333) = 53.13°.

Bearing of R from P: 030° + 53.13° = 83.13° ≈ 083°. ✓

L is due north of P: L = (0, y). L is due west of R: L = (29.78 - d, 3.59). So L = (0, 3.59). PL = 3.59 km.

This is a reasonable answer but not a nice number. Let me adjust to get a nicer answer.

Actually, let me just use the original numbers and redesign part (d) to avoid the contradiction.

Revised 15(d): A lighthouse is located at point L. L is the point on the north line from P that is closest to R. Calculate the distance RL.

With P at (0,0), R at (28.51, -9.34): The north line from P is the y-axis (x = 0). The closest point on x = 0 to R is L = (0, -9.34). RL = 28.51 km.

But this doesn't use "due north" in a meaningful way. Let me try a different approach.

Alternative 15(d): Calculate the distance of R from the north-south line through P.

With R at (28.51, -9.34), the distance from the north-south line through P (the y-axis) is |28.51| = 28.51 km.

$\boxed{RL = 28.5 \text{ km (3 s.f.)}}$

Marking notes for 15(d): M1 for correct interpretation of the geometry. A1 for correct distance.

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16.

(a) [3]

$\text{Area of triangle ABC} = \frac{1}{2}(AB)(BC)\sin(\angle ABC)$

$= \frac{1}{2}(200)(150)\sin 52°$

$= \frac{1}{2}(200)(150)(0.7880)$

$= 11820.3$

$\boxed{\text{Area} = 11800 \text{ m}^2 \text{ (3 s.f.)}}$

(b) [2]

Since D lies on AB and CD divides the area in half:

Area of triangle BCD = ½ × Area of triangle ABC = 5910.1 m²

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

$5910.1 = \frac{1}{2} \times BD \times 150 \times \sin 52°$

$5910.1 = \frac{1}{2} \times BD \times 150 \times 0.7880$

$5910.1 = 59.10 \times BD$

$BD = \frac{5910.1}{59.10} = 100$

$AD = AB - BD = 200 - 100 = 100$

$\boxed{AD = 100 \text{ m (3 s.f.)}}$

(c) [2]

Using the cosine rule in triangle BCD to find CD:

$CD^2 = BC^2 + BD^2 - 2(BC)(BD)\cos(\angle CBD)$

$CD^2 = 150^2 + 100^2 - 2(150)(100)\cos 52°$

$CD^2 = 22500 + 10000 - 30000(0.6157)$

$CD^2 = 32500 - 18470.9 = 14029.1$

$CD = \sqrt{14029.1} = 118.44$

$\boxed{CD = 118 \text{ m (3 s.f.)}}$

Marking notes for 16(c): M1 for correct cosine rule. A1 for correct length.

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Summary of Marks

Section	Marks
A (Q1–10)	20
B (Q11–14)	25
C (Q15–16)	15
Total	60