IEB mock Paper II (second paper)

(a) Answer: P$\hat{\mathrm{T}}$A = 42°

1. P$\hat{\mathrm{T}}$A = T$\hat{\mathrm{B}}$A = 42° (tan chord theorem)

(b) Answer: T$\hat{\mathrm{O}}$A = 84°

1. T$\hat{\mathrm{O}}$A = 2 × T$\hat{\mathrm{B}}$A = 84° (∠ at centre = 2 × ∠ at circumference)

(c) Answer: O$\hat{\mathrm{T}}$A = 48°

1. O$\hat{\mathrm{T}}$P = 90° (tan ⊥ radius)
2. ∴ O$\hat{\mathrm{T}}$A = 90° − P$\hat{\mathrm{T}}$A = 48°

(d) Answer: T$\hat{\mathrm{D}}$A = 138°

1. T$\hat{\mathrm{D}}$A = 180° − T$\hat{\mathrm{B}}$A = 138° (opp ∠s of cyclic quad)

(e) Answer: reflex T$\hat{\mathrm{O}}$A = 276°

1. reflex T$\hat{\mathrm{O}}$A = 360° − 84° = 276° (∠s round a pt)

(a) Answer: y = 10.873 + 7.708x

1. A = 10.873
2. B = 7.708

(b) Answer: r = 1.00

(c) Answer: a strong positive correlation

1. r = 1.00: r > 0 and |r| is at least 0.8

(d) Answer: y = 280.65

1. y = 10.873 + 7.708(35)

(e) Answer: No: x lies outside the data (extrapolation)

1. the data runs from x = 7 to x = 29; x = 35 lies outside it

(a) Answer: sin 324° = −k

1. sin 324° = sin(360° − 36°) = −sin 36° = −k

(b) Answer: tan 36° = k/√(1 − k^{^2})

1. cos^{^2} 36° = 1 − sin^{^2} 36° = 1 − k^{^2}
2. 36° is acute, so cos 36° > 0
3. tan 36° = (sin 36°)/(cos 36°) = k/√(1 − k^{^2})

(c) Answer: cos 72° = 1 − 2k^{^2}

1. cos 72° = 1 − 2sin^{^2} 36°
2. = 1 − 2k^{^2}

(d) Answer: sin 77° cos 41° − cos 77° sin 41° = k

1. sin 77° cos 41° − cos 77° sin 41° = sin(77° − 41°)
2. = sin 36° = k

(a) Answer: M(1; 3)

1. the diagonals of a square bisect each other: M is the midpoint of AC
2. M = ((−1 + 3)/2; (2 + 4)/2) = (1; 3)

(b) Answer: y = −2x + 5

1. m_{_AC} = (4 − 2)/(3 − (−1)) = 1/2
2. the diagonals of a square are perpendicular: m_{_BD} = −2
3. through M: 3 = (−2)(1) + c, so c = 5

(c) Answer: B(2; 1) and D(0; 5)

1. the diagonals of a square are equal and bisect each other: MB = MD = MA
2. MA^{^2} = (−1 − 1)^{^2} + (2 − 3)^{^2} = 5
3. on BD: (x − 1)^{^2} + (−2x + 2)^{^2} = 5
4. x = 2 or x = 0

(d) Answer: 10 square units

1. AC^{^2} = 4 × MA^{^2} = 20
2. the diagonals are equal and perpendicular: area = ½ × AC × BD = ½ × AC^{^2} = 10

(e) Answer: θ = 26.6°

1. tan θ = m_{_AC} = 1/2

(a) Answer: See the proof below.

1. Required to prove: B$\hat{\mathrm{A}}$T = A$\hat{\mathrm{C}}$B
2. Construction: Draw the diameter AD and join DB.
3. D$\hat{\mathrm{A}}$T = 90° (tan ⊥ radius)
4. A$\hat{\mathrm{B}}$D = 90° (∠ in semi-circle)
5. ∴ B$\hat{\mathrm{A}}$T = 90° − D$\hat{\mathrm{A}}$B
6. A$\hat{\mathrm{D}}$B = 90° − D$\hat{\mathrm{A}}$B (sum ∠s of Δ)
7. ∴ B$\hat{\mathrm{A}}$T = A$\hat{\mathrm{D}}$B
8. A$\hat{\mathrm{D}}$B = A$\hat{\mathrm{C}}$B (∠s in the same seg)
9. ∴ B$\hat{\mathrm{A}}$T = A$\hat{\mathrm{C}}$B

(b) Answer: P$\hat{\mathrm{A}}$D = A$\hat{\mathrm{E}}$D

1. P$\hat{\mathrm{A}}$B = A$\hat{\mathrm{C}}$B (tan chord theorem)
2. A$\hat{\mathrm{C}}$B = A$\hat{\mathrm{E}}$D (corresp ∠s; DE ∥ BC)
3. ∴ P$\hat{\mathrm{A}}$D = A$\hat{\mathrm{E}}$D

(c) Answer: PQ is a tangent to the circle through A, D and E

1. P$\hat{\mathrm{A}}$D = A$\hat{\mathrm{E}}$D (proved)
2. ∴ PQ is a tangent to the circle through A, D and E (converse tan chord theorem)

(d) Answer: AE = 6 units

1. AE/EC = AD/DB = 2 (line ∥ one side of Δ; DE ∥ BC)
2. ∴ AE = (4/6) × 9 = 6 units

(a) Answer: a = 3 and p = −60°

1. the greatest value of f is 3, so a = 3
2. cos(x + p) is greatest where x + p = 0°
3. 60° + p = 0°, so p = −60°

(b) Answer: −3 ≤ y ≤ 3

1. f runs from −3 to 3

(c) Answer: See the sketch below.

1. g(x) = 3 sin x
2. g: x-intercepts (−180°; 0), (0°; 0), (180°; 0)
3. g: y-intercept (0°; 0)
4. g: turning points (−90°; −3), (90°; 3)
5. g: end points (−180°; 0), (180°; 0)

(d) Answer: x = −105° or x = 75°

1. 3 cos(x − 60°) = 3 sin x
2. cos(x − 60°) = cos(90° − x)
3. x − 60° = 90° − x + k·360°, so x = 75° + k·180°
4. or x − 60° = −(90° − x) + k·360°, which gives −60° = −90° + k·360°: no solution
5. In [−180°; 180°]: x = −105° or x = 75°

(e) Answer: −30° < x < 0° or 150° < x < 180°

1. f and g have opposite signs, read from the sketch

(a) Answer: FG : GC = 5 : 4

1. In ΔCBF: DG ∥ BF
2. ∴ FG : GC = 5 : 4 (line ∥ one side of Δ; DG ∥ BF)

(b) Answer: AF = FG

1. In ΔADG: AE = ED and EF ∥ DG
2. ∴ AF = FG (line through midpt ∥ to second side)

(c) Answer: AF : FC = 5 : 9

1. AF = FG = 5k and GC = 4k (proved)
2. ∴ AF : FC = 5 : 9

(d) Answer: 5 : 9

1. area ΔABF : area ΔCBF = 5 : 9 (same height)

(a) Answer: M(0; 4); r = √17

1. x^{^2} + (y − 4)^{^2} = 1 + 16
2. x^{^2} + (y − 4)^{^2} = 17
3. M(0; 4) and r = √(17) = √17

(b) Answer: See the working below.

1. (1)^{^2} + (8)^{^2} + 0(1) − 8(8) − 1 = 0
2. so P lies on the circle

(c) Answer: y = −(1/4)x + 33/4

1. m_{_MP} = (8 − 4)/(1 − 0) = 4
2. the tangent is perpendicular to the radius: m = −1/4
3. 8 = (−1/4)(1) + c, so c = 33/4

(d) Answer: QT = 9 units

1. QM^{^2} = (−7 − 0)^{^2} + (11 − 4)^{^2} = 98
2. MT ⊥ QT (radius ⊥ tangent), so QT^{^2} = QM^{^2} − r^{^2}
3. QT^{^2} = 98 − 17 = 81
4. QT = 9

(a) Answer: See the working below.

1. m_{_AB} = (−4 − 2)/(−4 − (−2)) = 3
2. m_{_BC} = (−6 − (−4))/(2 − (−4)) = −1/3
3. m_{_AB} × m_{_BC} = (3)(−1/3) = −1, so AB ⊥ BC

(b) Answer: M(0; −2)

1. A$\hat{\mathrm{B}}$C = 90°, so AC subtends a right angle at B: AC is a diameter (converse ∠ in semi-circle)
2. M is the midpoint of AC: ((−2 + 2)/2; (2 + (−6))/2) = (0; −2)

(c) Answer: x^{^2} + (y + 2)^{^2} = 20

1. r^{^2} = MA^{^2} = (−2 − 0)^{^2} + (2 − (−2))^{^2} = 20

(d) Answer: y = (1/2)x − 7

1. m_{_MC} = −2
2. the tangent is perpendicular to the radius MC: m = 1/2
3. −6 = (1/2)(2) + c, so c = −7

(e) Answer: 20 square units

1. AB = 2√10 and BC = 2√10
2. area = ½ × AB × BC (the right angle is at B) = 20

(a) Answer: CDHE is a cyclic quadrilateral

1. H$\hat{\mathrm{D}}$C = 90° (given)
2. H$\hat{\mathrm{E}}$C = 90° (given)
3. ∴ H$\hat{\mathrm{D}}$C + H$\hat{\mathrm{E}}$C = 180°
4. ∴ CDHE is a cyclic quad (opp ∠s of quad supp)

(b) Answer: ABDE is a cyclic quadrilateral

1. A$\hat{\mathrm{D}}$B = A$\hat{\mathrm{E}}$B = 90° (given)
2. ∴ ABDE is a cyclic quad (line subtends equal ∠s)

(c) Answer: D$\hat{\mathrm{A}}$C = 90° − x

1. A$\hat{\mathrm{D}}$C = 90° (given)
2. ∴ D$\hat{\mathrm{A}}$C = 90° − x (sum ∠s of Δ)

(d) Answer: ${\hat{\mathrm{D}}}_1$ = B$\hat{\mathrm{A}}$C

1. ${\hat{\mathrm{D}}}_1$ = B$\hat{\mathrm{A}}$E (ext ∠ of cyclic quad)

(e) Answer: ΔCDE ||| ΔCAB

1. In ΔCDE and ΔCAB:
2. D$\hat{\mathrm{C}}$E = A$\hat{\mathrm{C}}$B (common)
3. ${\hat{\mathrm{D}}}_1$ = C$\hat{\mathrm{A}}$B (proved)
4. ∴ ΔCDE ||| ΔCAB (∠∠∠)

(f) Answer: CB = 30 units

1. CD/CA = CE/CB (||| Δs)
2. ∴ CB = (18 · 25)/15 = 30 units

(a) Answer: 21.08

1. midpoints: 5, 15, 25, 35, 45
2. Σ f·x = 1075 and n = 51
3. mean = 1075/51 = 21.08

(b) Answer: 10 ≤ x < 20

1. the highest frequency is 16

(c) Answer: See the ogive below.

1. cumulative frequencies: 10, 26, 41, 45, 51
2. plot each against the upper limit of its class, starting at (0; 0)

(d) Answer: about 19.7

1. the median is at a cumulative frequency of 51/2 = 25.5
2. read across to the ogive and down: about 19.7

(e) Answer: about 16.5

1. Q1 at 12.75: about 11.7; Q3 at 38.25: about 28.2
2. IQR = 28.2 − 11.7 = 16.5

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

1. mean 21.08 and median about 19.7
2. mean > median

(a) Answer: See the working below.

1. AT makes an angle α and the road an angle θ with the horizontal: P$\hat{\mathrm{A}}$T = α − θ
2. the horizontal at P makes 90° with PT, and PA lies θ below it: A$\hat{\mathrm{P}}$T = 90° + θ
3. A$\hat{\mathrm{T}}$P = 180° − (α − θ) − (90° + θ) = 90° − α (sum ∠s of Δ)

(b) Answer: See the working below.

1. In ΔAPT: PT/(sin P$\hat{\mathrm{A}}$T) = AP/(sin A$\hat{\mathrm{T}}$P) (sine rule)
2. PT = (d sin(α − θ))/(sin(90° − α))
3. sin(90° − α) = cos α, so PT = (d sin(α − θ))/(cos α)

(c) Answer: PT = 9.81 m

1. PT = (26 sin 20°)/(cos 25°)
2. = 9.81 m

(d) Answer: BT = 47.86 m

1. AT = (d sin(90° + θ))/(sin(90° − α)) = (26 cos 5°)/(cos 25°) = 28.58 m (sine rule)
2. B$\hat{\mathrm{A}}$T = 180° − 20° = 160° (∠s on a str line)
3. BT^{^2} = 20^{^2} + 28.58^{^2} − 2(20)(28.58) cos 160° (cosine rule)

(e) Answer: 16.78°

1. (sin A$\hat{\mathrm{B}}$T)/AT = (sin B$\hat{\mathrm{A}}$T)/BT (sine rule)
2. A$\hat{\mathrm{B}}$T = 11.78°
3. the road rises at 5°: the angle of elevation is 11.78° + 5° = 16.78°