Secondary 3 Elementary Mathematics Semestral Assessment 2 (End of Year) Paper 5

TuitionGoWhere Practice Paper – Elementary Mathematics Secondary 3

SA2 – Version 5: Answer Key and Marking Scheme

Total Marks: 60

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Section A: Short Answer Questions (30 marks)

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Question 1

(a) $AC = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17$ cm [1]

(b) $\sin \angle BAC = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{BC}{AC} = \dfrac{15}{17}$ [1]

(c) $\angle BAC = \sin^{-1}\left(\dfrac{15}{17}\right) = 61.927...^\circ \approx 61.9^\circ$ [2]

Marking: 1 mark for correct ratio, 1 mark for correct angle to 1 d.p.

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Question 2

(a) Using cosine rule:
$QR^2 = 12^2 + 9^2 - 2(12)(9)\cos 110^\circ$
$QR^2 = 144 + 81 - 216 \times (-0.34202...)$
$QR^2 = 225 + 73.876...$
$QR^2 = 298.876...$
$QR = 17.288... \approx 17.3$ cm [2]

Marking: 1 mark for correct substitution, 1 mark for correct answer.

(b) Area $= \dfrac{1}{2}ab\sin C = \dfrac{1}{2} \times 12 \times 9 \times \sin 110^\circ$
$= 54 \times 0.93969...$
$= 50.743... \approx 50.7$ cm² [2]

Marking: 1 mark for correct formula, 1 mark for correct answer.

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Question 3

(a) $\angle ACB = \dfrac{1}{2}\angle AOB = \dfrac{1}{2} \times 130^\circ = 65^\circ$
(Angle at centre = 2 × angle at circumference) [1]

(b) $\angle ADB = \angle ACB = 65^\circ$
(Angles in the same segment are equal) [1]

(c) $\angle CAD = \angle CBD = 42^\circ$ (Angles in the same segment)
In $\triangle ACD$: $\angle ACD = 180^\circ - 65^\circ - 35^\circ = 80^\circ$
$\angle CAD = 180^\circ - 80^\circ - 35^\circ = 65^\circ$
Alternatively: $\angle CAD = \angle CBD$ (angles in same segment)
$\angle CAD = 180^\circ - 65^\circ - 35^\circ - 42^\circ = 38^\circ$
Wait – need to check diagram logic.
$\angle CAD = \angle CBD = 42^\circ$ (angles subtended by arc $CD$) [2]

Marking: 1 mark for identifying correct angle relationship, 1 mark for correct answer.

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Question 4

(a) Diagram showing right-angled triangle $PTQ$ with $TQ$ vertical, $PQ = 120$ m, $\angle TPQ = 28^\circ$. [1]

(b) $\tan 28^\circ = \dfrac{TQ}{120}$
$TQ = 120 \times \tan 28^\circ = 120 \times 0.53170... = 63.804... \approx 63.8$ m [2]

Marking: 1 mark for correct trig ratio, 1 mark for correct answer.

(c) $PR = \sqrt{120^2 + 50^2} = \sqrt{14400 + 2500} = \sqrt{16900} = 130$ m
Angle of elevation from $R$: $\tan \theta = \dfrac{63.8}{130}$
$\theta = \tan^{-1}\left(\dfrac{63.8}{130}\right) = \tan^{-1}(0.49076...) = 26.13...^\circ \approx 26.1^\circ$ [3]

Marking: 1 mark for finding $PR$, 1 mark for correct trig ratio, 1 mark for correct answer.

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Question 5

(a) Diagram showing $A$, $B$, $C$ with bearings $055^\circ$ and $145^\circ$, distances 8 km and 12 km. [1]

(b) Angle $ABC = 145^\circ - 55^\circ = 90^\circ$ (difference in bearings)
$AC = \sqrt{8^2 + 12^2} = \sqrt{64 + 144} = \sqrt{208} = 14.422... \approx 14.4$ km [2]

Marking: 1 mark for identifying right angle, 1 mark for correct answer.

(c) $\tan \angle BAC = \dfrac{12}{8} = 1.5$
$\angle BAC = \tan^{-1}(1.5) = 56.309...^\circ$
Bearing of $C$ from $A = 055^\circ + 56.3^\circ = 111.3^\circ$ [2]

Marking: 1 mark for finding angle $BAC$, 1 mark for correct bearing.

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Question 6

(a) Let cuboid have $A$ at origin, edges along axes: $A(0,0,0)$, $F(6,0,10)$, $G(6,8,10)$, $M(6,4,10)$.
$AM = \sqrt{6^2 + 4^2 + 10^2} = \sqrt{36 + 16 + 100} = \sqrt{152} = 12.328... \approx 12.3$ cm [2]

Marking: 1 mark for correct coordinates or Pythagoras steps, 1 mark for correct answer.

(b) $H(0,8,10)$. $AH = \sqrt{0^2 + 8^2 + 10^2} = \sqrt{164} = 12.806...$ cm
$MH = \sqrt{6^2 + 4^2 + 0^2} = \sqrt{52} = 7.211...$ cm
Using cosine rule in $\triangle AMH$:
$\cos \angle AMH = \dfrac{AM^2 + MH^2 - AH^2}{2 \times AM \times MH}$
$= \dfrac{152 + 52 - 164}{2 \times 12.328 \times 7.211}$
$= \dfrac{40}{177.78...} = 0.22498...$
$\angle AMH = \cos^{-1}(0.22498...) = 77.00...^\circ \approx 77.0^\circ$ [2]

Marking: 1 mark for finding all three sides, 1 mark for correct angle.

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Section B: Structured Questions (30 marks)

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Question 7

(a) Using cosine rule:
$XZ^2 = 7^2 + 9^2 - 2(7)(9)\cos 65^\circ$
$= 49 + 81 - 126 \times 0.42261...$
$= 130 - 53.249...$
$= 76.750...$
$XZ = 8.760... \approx 8.76$ cm [2]

Marking: 1 mark for correct substitution, 1 mark for correct answer.

(b) Using sine rule:
$\dfrac{\sin \angle YXZ}{9} = \dfrac{\sin 65^\circ}{8.76}$
$\sin \angle YXZ = \dfrac{9 \times \sin 65^\circ}{8.76} = \dfrac{9 \times 0.90630...}{8.76} = \dfrac{8.156...}{8.76} = 0.9311...$
$\angle YXZ = \sin^{-1}(0.9311...) = 68.58...^\circ \approx 68.6^\circ$ [2]

Marking: 1 mark for correct sine rule setup, 1 mark for correct answer.

(c) Area $= \dfrac{1}{2} \times 7 \times 9 \times \sin 65^\circ$
$= 31.5 \times 0.90630... = 28.548... \approx 28.5$ cm² [2]

Marking: 1 mark for correct formula, 1 mark for correct answer.

(d) $XW = XY \times \sin \angle XYZ = 7 \times \sin 65^\circ = 7 \times 0.90630... = 6.344... \approx 6.34$ cm
Alternatively: $XW = \dfrac{2 \times \text{Area}}{YZ} = \dfrac{2 \times 28.55}{9} = 6.344... \approx 6.34$ cm [2]

Marking: 1 mark for correct method, 1 mark for correct answer.

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Question 8

(a) $\angle ACB = 90^\circ$ because the angle in a semicircle is a right angle (angle subtended by diameter $AB$). [1]

(b) $\angle ACD = \angle ABD$ (angles in the same segment)
$\angle ABD = 180^\circ - 90^\circ - 42^\circ = 48^\circ$ (in $\triangle ABD$, angle in semicircle at $D$)
Wait – need to reconsider.
$\angle ACD = \angle ABD$ (angles subtended by arc $AD$)
In $\triangle ABD$: $\angle ADB = 90^\circ$ (angle in semicircle)
$\angle ABD = 180^\circ - 90^\circ - 28^\circ = 62^\circ$
So $\angle ACD = 62^\circ$ [2]

Marking: 1 mark for identifying correct angle relationship, 1 mark for correct answer.

(c) $\angle AOD = 2 \times \angle ACD = 2 \times 62^\circ = 124^\circ$
(Angle at centre = 2 × angle at circumference) [2]

Marking: 1 mark for correct relationship, 1 mark for correct answer.

(d) $\angle CAB = 90^\circ - 28^\circ = 62^\circ$ (in right-angled $\triangle ACB$)
$\angle ACD = 62^\circ$ (from part b)
Since $\angle CAB = \angle ACD$, and these are alternate angles, $CD \parallel AB$. [3]

Marking: 1 mark for finding $\angle CAB$, 1 mark for equating to $\angle ACD$, 1 mark for conclusion with reason.

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Question 9

(a) Using cosine rule in $\triangle ABC$:
$AC^2 = 6^2 + 8^2 - 2(6)(8)\cos 100^\circ$
$= 36 + 64 - 96 \times (-0.17364...)$
$= 100 + 16.670...$
$= 116.670...$
$AC = 10.801... \approx 10.8$ cm [2]

Marking: 1 mark for correct substitution, 1 mark for correct answer.

(b) $\angle ADC = 180^\circ - 100^\circ = 80^\circ$
(Opposite angles of a cyclic quadrilateral sum to $180^\circ$) [1]

(c) Using cosine rule in $\triangle ADC$:
$\cos \angle DAC = \dfrac{7^2 + 10.8^2 - 5^2}{2 \times 7 \times 10.8}$
$= \dfrac{49 + 116.64 - 25}{151.2} = \dfrac{140.64}{151.2} = 0.9301...$
$\angle DAC = \cos^{-1}(0.9301...) = 21.56...^\circ$
$\angle BAD = \angle BAC + \angle CAD$
In $\triangle ABC$: $\cos \angle BAC = \dfrac{6^2 + 10.8^2 - 8^2}{2 \times 6 \times 10.8} = \dfrac{36 + 116.64 - 64}{129.6} = \dfrac{88.64}{129.6} = 0.6839...$
$\angle BAC = \cos^{-1}(0.6839...) = 46.85...^\circ$
$\angle BAD = 46.85^\circ + 21.56^\circ = 68.41...^\circ \approx 68.4^\circ$ [2]

Marking: 1 mark for finding one component angle, 1 mark for correct total.

(d) In $\triangle ABE$: $\angle AEB = 180^\circ - \angle BAE - \angle ABE$
$\angle ABE = \angle ABD$ (same angle)
Using sine rule in $\triangle ABD$: $\dfrac{\sin \angle ABD}{7} = \dfrac{\sin 80^\circ}{BD}$
Need $BD$ first. Using cosine rule in $\triangle BCD$:
$BD^2 = 8^2 + 5^2 - 2(8)(5)\cos \angle BCD$
$\angle BCD = 180^\circ - \angle BAD = 180^\circ - 68.4^\circ = 111.6^\circ$
$BD^2 = 64 + 25 - 80 \times (-0.3681...) = 89 + 29.45 = 118.45$
$BD = 10.88...$ cm
$\sin \angle ABD = \dfrac{7 \times \sin 80^\circ}{10.88} = \dfrac{7 \times 0.98480...}{10.88} = 0.6335...$
$\angle ABD = 39.30...^\circ$
$\angle BEC = \angle AED$ (vertically opposite)
In $\triangle AED$: $\angle AED = 180^\circ - 21.56^\circ - (80^\circ - 39.30^\circ) = 180^\circ - 21.56^\circ - 40.70^\circ = 117.74^\circ$
$\angle BEC = 117.7^\circ$ [3]

Marking: 1 mark for finding $BD$, 1 mark for finding relevant angles, 1 mark for correct answer.

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Question 10

(a) $\tan 35^\circ = \dfrac{15}{FP}$
$FP = \dfrac{15}{\tan 35^\circ} = \dfrac{15}{0.70020...} = 21.422... \approx 21.4$ m [2]

Marking: 1 mark for correct trig ratio, 1 mark for correct answer.

(b) $\tan 50^\circ = \dfrac{15}{FQ}$
$FQ = \dfrac{15}{\tan 50^\circ} = \dfrac{15}{1.19175...} = 12.586... \approx 12.6$ m [2]

Marking: 1 mark for correct trig ratio, 1 mark for correct answer.

(c) $PQ = FP + FQ = 21.42 + 12.59 = 34.01 \approx 34.0$ m [1]

(d) Midpoint $M$ of $PQ$: $FM = FP - \dfrac{PQ}{2} = 21.42 - 17.005 = 4.415$ m
$\tan \theta = \dfrac{15}{4.415} = 3.397...$
$\theta = \tan^{-1}(3.397...) = 73.60...^\circ \approx 73.6^\circ$ [3]

Marking: 1 mark for finding $FM$, 1 mark for correct trig ratio, 1 mark for correct answer.

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END OF ANSWER KEY