IEB mock Paper II

A complete 150-mark practice paper: 12 questions in 57 parts, 28 topics, Section A 75 marks and Section B 75. Every question was generated for this page and every answer was independently verified.

The split between the sections, and how much of the paper each topic is worth, were measured from 282 real IEB questions — not guessed. None of those questions appears here.

All IEB practice · Sit this paper under time · Afrikaans

The paper

Section A[75 marks]

1[9 marks]

In the diagram, ΔABC is drawn.

D is the mid-point of AB.
E lies on AC with DE ∥ BC.
H lies on BC with BH : HC = 2 : 3, and AH cuts DE at K.
BC = 10 units.

(a)
Prove that E is the mid-point of AC.
(2)
(b)
Calculate the length of DE.
(2)
(c)
Calculate the length of DK.
(3)
(d)
Determine DK : KE.
(2)

2[10 marks]

Given:

sin A = −7/25 where 180° < A < 270°
sin B = −3/5 where 270° < B < 360°

(a)
Determine, without the use of a calculator, the value of cos A.
(2)
(b)
Hence determine, without the use of a calculator, the value of cos(A + B).
(5)
(c)
Determine the value of sin 2B.
(3)

3[13 marks]

In the diagram below (answers correct to two decimal places):

ABC is a triangular field on horizontal ground, and D is a point on AC.
EB is a vertical lamp post at B.
AB = 110 m and DC = 50 m.
B $\hat{A}$ D = 50° and B $\hat{D}$ A = 70°.

(a)
Calculate the length of BD.
(3)
(b)
Calculate the length of BC.
(4)
(c)
Calculate the area of ΔBDC.
(3)
(d)
The angle of elevation of E from D is 28°. Calculate the height of the lamp post.
(3)

4[14 marks]

In the diagram, A(−5; −1), B(0; 4), D(2; −2) and E(10; 4) are given. AE and DB intersect at C(1; k).

(a)
Calculate the length of AB, in simplest surd form.
(2)
(b)
Determine the equation of the line DB.
(3)
(c)
Determine the value of k.
(2)
(d)
Prove that AE ⊥ DB.
(3)
(e)
Determine the size of A $\hat{B}$ D, correct to one decimal place.
(4)

5[14 marks]

In the diagram, ΔABC with D on AB and E on AC, and DE ∥ BC.

(a)
Use the diagram to prove the theorem which states that a line parallel to one side of a triangle divides the other two sides proportionally.
(4)
In the diagram below, ΔABC is drawn. E is a point on AC and D is a point on BC with AB ∥ ED. F is a point on ED and G is a point on EC with FG ∥ BC. AE = 9 units, BD = 8 units and DC = 4 units. EF : FD = 3 : 2.
(b)
Calculate the length of EC.
(3)
(c)
Calculate the length of EG.
(3)
(d)
Calculate the length of FG.
(4)

6[15 marks]

The frequency polygon shows the mass in kg of 52 parcels, grouped in the classes 0 ≤ x < 5; 5 ≤ x < 10; 10 ≤ x < 15; 15 ≤ x < 20; 20 ≤ x < 25.

(a)
Estimate the mean of the data.
(3)
(b)
Write down the modal class.
(1)
(c)
Draw an ogive (cumulative frequency curve) for the data.
(4)
(d)
Use your ogive to estimate the median.
(2)
(e)
Use your ogive to estimate the interquartile range.
(3)
(f)
Is the data skewed? Explain, using the mean and the median.
(2)

Section B[75 marks]

7[8 marks]

In the diagram, ΔABC is drawn.

E and F are points on AC and AB respectively, with AE : EC = 5 : 2 and AF : FB = 3 : 5.
BC produced meets FE produced at D.
G is a point on FB so that FD ∥ GC.

(a)
Determine AF : FG.
(2)
(b)
Hence determine FG : GB.
(3)
(c)
Calculate BC : CD.
(3)

8[11 marks]

Given: sin 38° = k. Determine, without the use of a calculator, each of the following in terms of k.

(a)
sin 218°
(2)
(b)
tan 38°
(3)
(c)
cos 76°
(3)
(d)
sin 56° cos 18° − cos 56° sin 18°
(3)

9[12 marks]

Circle P has centre P(−5; −3) and touches the x-axis. Circle Q has the equation (x + 17)^{^2} + (y + 12)^{^2} = 144.

(a)
Write down the equation of circle P.
(2)
(b)
Calculate the distance PQ between the centres.
(2)
(c)
Show that the circles touch externally.
(2)
(d)
Determine the coordinates of T, the point where the circles touch.
(3)
(e)
Determine the equation of the common tangent to the circles at T.
(3)

10[14 marks]

In the diagram, AB is a diameter of the circle with centre M, where A(−5; 6) and B(5; 4).

(a)
Determine the coordinates of M, the centre of the circle.
(2)
(b)
Determine the equation of the circle.
(3)
(c)
Show that P(−1; 0) lies on the circle.
(2)
(d)
Determine the equation of the tangent to the circle at A.
(4)
(e)
Determine the y-intercepts of the circle, in simplest surd form where needed.
(3)

11[14 marks]

In the diagram, O is the centre of the circle and OM ⊥ AB, with M on AB. O $\hat{A}$ B = 42°.

(a)
Determine, with reasons, the size of O $\hat{M}$ A.
(2)
(b)
Determine, with reasons, the size of O $\hat{M}$ B.
(2)
(c)
Determine, with reasons, the size of A $\hat{O}$ M.
(2)
In the diagram below, AB is a diameter of the circle with centre O, and P and Q lie on the circle. P $\hat{A}$ B = 73°.
(d)
Determine, with reasons, the size of A $\hat{P}$ B.
(2)
(e)
Determine, with reasons, the size of B $\hat{Q}$ A.
(2)
(f)
Determine, with reasons, the size of B $\hat{Q}$ P.
(2)
(g)
Determine, with reasons, the size of P $\hat{B}$ O.
(2)

12[16 marks]

The graph of f(x) = cos bx is sketched below for x ∈ [−90°; 270°].

(a)
Determine the value of b.
(1)
(b)
Write down the period of f.
(1)
(c)
Sketch the graph of g(x) = −cos x on the same set of axes as f. Label all intercepts with the axes, turning points and end points.
(4)
(d)
Write down the amplitude of g.
(1)
(e)
Determine the values of x ∈ [−90°; 270°] for which f(x) = g(x). Show all calculations.
(5)
(f)
Use your graph to determine the values of x ∈ [−90°; 270°] for which f(x) > g(x).
(2)
(g)
If g is shifted 1 unit up and 30° to the right, write down the new equation of g in the form y = …
(2)

Memorandum

Every answer below was confirmed by an independent verifier at the moment the question was made.

(a) Answer: See the proof below.

AD = DB (given)
DE ∥ BC (given)
∴ AE = EC (line from midpoint ∥ to one side)

(b) Answer: DE = 5 units

DE = ½BC (Midpt Theorem)
DE = ½ × 10 = 5

(c) Answer: DK = 2 units

BH = 2/5 × 10 = 4
In ΔABH: AD = DB and DK ∥ BH (given)
∴ AK = KH (line from midpoint ∥ to one side)
DK = ½BH = 2 (Midpt Theorem)

(d) Answer: DK : KE = 2 : 3

In ΔAHC: KE = ½HC (Midpt Theorem)
∴ DK : KE = ½BH : ½HC = 2 : 3

The midpoint theorem

(a) Answer: cos A = −24/25

A lies in the third quadrant: y = −7, r = 25
x^{^2} = 25^{^2} − (−7)^{^2} = 576, so x = −24
sin A = −7/25 and cos A = −24/25

(b) Answer: cos(A + B) = −117/125

B lies in the fourth quadrant: y = −3, r = 5
x^{^2} = 5^{^2} − (−3)^{^2} = 16, so x = 4
sin B = −3/5 and cos B = 4/5
cos(A + B) = cos A cos B − sin A sin B
= (−24/25)(4/5) − (−7/25)(−3/5)
= −117/125

(c) Answer: sin 2B = −24/25

sin 2B = 2 sin B cos B
= 2(−3/5)(4/5)
= −24/25

Identities, reduction and compound angles

(a) Answer: BD = 89.67 m

In ΔABD: (BD)/(sin B $\hat{A}$ D) = (AB)/(sin B $\hat{D}$ A)
BD = (110 sin 50°)/(sin 70°)
= 89.67 m

(b) Answer: BC = 116.65 m

B $\hat{D}$ C = 180° − 70° = 110° (∠s on a str line)
BC^{^2} = BD^{^2} + DC^{^2} − 2(BD)(DC) cos B $\hat{D}$ C
= 89.67^{^2} + 50^{^2} − 2(89.67)(50) cos 110°
BC = 116.65 m

(c) Answer: area = 2106.62 m²

Area = (1/2)(BD)(DC) sin B $\hat{D}$ C
= (1/2)(89.67)(50) sin 110°
= 2106.62 m²

(d) Answer: EB = 47.68 m

In ΔEBD, E $\hat{B}$ D = 90°: tan 28° = (EB)/(BD)
EB = 89.67 tan 28°
= 47.68 m

Sine, cosine and area rules (2D)

(a) Answer: AB = 5√2 units

AB = √((0 − (−5))^{^2} + (4 − (−1))^{^2})
= √(50) = 5√2

(b) Answer: y = −3x + 4

m_{_DB} = (4 − (−2))/(0 − (2)) = −3
−2 = (−3)(2) + c, so c = 4
y = −3x + 4

(c) Answer: k = 1

C lies on DB: k = (−3)(1) + (4)
k = 1

(d) Answer: See the proof below.

m_{_AE} = (4 − (−1))/(10 − (−5)) = 1/3
m_{_AE} × m_{_DB} = (1/3)(−3) = −1
∴ AE ⊥ DB

(e) Answer: A $\hat{B}$ D = 63.4°

the inclination of AB is 45.0° and of DB is 108.4°
the lines make 63.4°
A $\hat{B}$ D = 63.4°

The distance between two points

(a) Answer: See the proof below.

Given: ΔABC with D on AB and E on AC, and DE ∥ BC.
To prove: AD/DB = AE/EC
Construction: Join BE and CD. Draw EX ⊥ AB and DY ⊥ AC.
area ΔADE / area ΔBDE = AD/DB (same height)
area ΔADE / area ΔCDE = AE/EC (same height)
area ΔBDE = area ΔCDE (same base; same height)
∴ AD/DB = AE/EC

(b) Answer: EC = 4.5 units

CE/EA = CD/DB (line ∥ one side of Δ; ED ∥ AB)
EC/9 = 4/8
EC = 4.5

(c) Answer: EG = 2.7 units

EG/GC = EF/FD = 3/2 (line ∥ one side of Δ; FG ∥ DC)
EG = 3/5 × EC = 3/5 × 4.5
EG = 2.7

(d) Answer: FG = 2.4 units

In ΔEFG and ΔEDC:
$\hat{E}$ = $\hat{E}$ (common)
E $\hat{F}$ G = C $\hat{D}$ E (corresp ∠s equal; FG ∥ DC)
∴ ΔEFG ||| ΔEDC (∠∠∠)
FG/DC = EF/ED = 3/5 (∥∥∥ Δs)
FG = 3/5 × 4 = 2.4

Proving the theorem itself, not using it

(a) Answer: 14.13

midpoints: 2.5, 7.5, 12.5, 17.5, 22.5
Σ f·x = 735 and n = 52
mean = 735/52 = 14.13

(b) Answer: 20 ≤ x < 25

the highest frequency is 18

(c) Answer: See the ogive below.

cumulative frequencies: 5, 18, 30, 34, 52
plot each against the upper limit of its class, starting at (0; 0)

(d) Answer: about 13.3

the median is at a cumulative frequency of 52/2 = 26
read across to the ogive and down: about 13.3

(e) Answer: about 13.3

Q1 at 13: about 8.1; Q3 at 39: about 21.4
IQR = 21.4 − 8.1 = 13.3

(f) Answer: Skewed to the right (positively)

mean 14.13 and median about 13.3
mean > median

The estimated mean of grouped data

(a) Answer: AF : FG = 5 : 2

AF/FG = AE/EC = 5/2 (line ∥ one side of Δ; FE ∥ GC)
AF : FG = 5 : 2

(b) Answer: FG : GB = 6 : 19

Let AF = 15k and FB = 25k (given)
FG = 2/5 × 15k = 6k
GB = FB − FG = 25k − 6k = 19k
FG : GB = 6 : 19

(c) Answer: BC : CD = 19 : 6

BC/CD = BG/GF (line ∥ one side of Δ; GC ∥ FD)
BC/CD = 19k/6k
BC : CD = 19 : 6

The proportion theorem

(a) Answer: sin 218° = −k

sin 218° = sin(180° + 38°) = −sin 38° = −k

(b) Answer: tan 38° = k/√(1 − k^{^2})

cos^{^2} 38° = 1 − sin^{^2} 38° = 1 − k^{^2}
38° is acute, so cos 38° > 0
tan 38° = (sin 38°)/(cos 38°) = k/√(1 − k^{^2})

(c) Answer: cos 76° = 1 − 2k^{^2}

cos 76° = 1 − 2sin^{^2} 38°
= 1 − 2k^{^2}

(d) Answer: sin 56° cos 18° − cos 56° sin 18° = k

sin 56° cos 18° − cos 56° sin 18° = sin(56° − 18°)
= sin 38° = k

Identities, reduction and compound angles

(a) Answer: (x + 5)^{^2} + (y + 3)^{^2} = 9

it touches the x-axis, so r = |−3| = 3
(x + 5)^{^2} + (y + 3)^{^2} = 9

(b) Answer: PQ = 15 units

PQ = √((−17 − (−5))^{^2} + (−12 − (−3))^{^2})
= 15

(c) Answer: See the working below.

r_{_P} + r_{_Q} = 3 + 12 = 15 = PQ
∴ the circles touch externally

(d) Answer: T(−37/5; −24/5)

T lies on PQ, 3 units from P: 3/15 of the way to Q
(−5 + (3/15)(−12); −3 + (3/15)(−9)) = (−37/5; −24/5)

(e) Answer: y = −(4/3)x − 44/3

m_{_PQ} = 3/4, so the tangent has m = −4/3
−24/5 = (−4/3)(−37/5) + c, so c = −44/3
y = −(4/3)x − 44/3

The equation of a circle

(a) Answer: M(0; 5)

M is the midpoint of the diameter AB
((−5 + (5))/2; (6 + (4))/2) = (0; 5)

(b) Answer: x^{^2} + (y − 5)^{^2} = 26

r^{^2} = MA^{^2} = (−5 − (0))^{^2} + (6 − (5))^{^2} = 26
x^{^2} + (y − 5)^{^2} = 26

(c) Answer: See the working below.

(−1 − (0))^{^2} + (0 − (5))^{^2} = 26 = r^{^2}
∴ P lies on the circle

(d) Answer: y = 5x + 31

m_{_MA} = −1/5
the tangent is perpendicular to the radius MA: m = 5
6 = (5)(−5) + c, so c = 31
y = 5x + 31

(e) Answer: y = 5 ± √26

x = 0: 0 + (y − (5))^{^2} = 26
(y − (5))^{^2} = 26
y = 5 ± √26

The midpoint of a line segment

(a) Answer: O $\hat{M}$ A = 90°

O $\hat{M}$ A = 90° (line from centre ⊥ to chord)

(b) Answer: O $\hat{M}$ B = 90°

O $\hat{M}$ B = 90° (line from centre ⊥ to chord)

(c) Answer: A $\hat{O}$ M = 48°

A $\hat{O}$ M = 48° (sum ∠s of Δ)

(d) Answer: A $\hat{P}$ B = 90°

A $\hat{P}$ B = 90° (∠s in semi-circle)

(e) Answer: B $\hat{Q}$ A = 90°

B $\hat{Q}$ A = 90° (∠s in semi-circle)

(f) Answer: B $\hat{Q}$ P = 73°

B $\hat{Q}$ P = 73° (∠s in the same seg)

(g) Answer: P $\hat{B}$ O = 17°

P $\hat{B}$ O = 17° (sum ∠s of Δ)

Chords, and the line from the centre

(a) Answer: b = 2

cos bx has period 360°/b
one full wave of f takes 180°
360°/b = 180°, so b = 2

(b) Answer: 180°

f(x) = cos 2x repeats every 180°

(c) Answer: See the sketch below.

g(x) = −cos x
g: x-intercepts (−90°; 0), (90°; 0), (270°; 0)
g: y-intercept (0°; −1)
g: turning points (0°; −1), (180°; 1)
g: end points (−90°; 0), (270°; 0)

(d) Answer: 1

the amplitude of a·cos x is |a| = 1

(e) Answer: x = −60° or x = 60° or x = 180°

cos 2x = −cos x
2 cos^{^2} x − 1 = −cos x
2 cos^{^2} x + cos x − 1 = 0
(2 cos x − 1)(cos x + 1) = 0
cos x = 1/2 or cos x = −1
x = ±60° + k·360° or x = 180° + k·360°
In [−90°; 270°]: x = −60° or x = 60° or x = 180°

(f) Answer: −60° < x < 60°

f lies above g, read from the sketch
−60° < x < 60°

(g) Answer: y = −cos(x − 30°) + 1

30° to the right: x becomes x − 30°
up 1: add 1
y = −cos(x − 30°) + 1

Trigonometric graphs