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

TuitionGoWhere Practice Paper - Elementary Mathematics Secondary 3 (SA2 Version 2) - Answer Key

Total Marks: 60

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Section A [30 marks]

1

(a) $\tan x = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AB} = \frac{8}{15}$

(b) $x = \tan^{-1}\left(\frac{8}{15}\right) = 28.072...^\circ = 28.1^\circ$ (1 d.p.)

Marks: (a) 1, (b) 1
Note: Identify opposite and adjacent correctly relative to angle $x$. Angle $x$ is at $A$, so opposite is $BC = 8$, adjacent is $AB = 15$.

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2

(a) $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{2.5}{6.5} = \frac{5}{13}$
$\theta = \cos^{-1}\left(\frac{5}{13}\right) = 67.380...^\circ = 67.4^\circ$ (1 d.p.)

(b) Height $h = \sqrt{6.5^2 - 2.5^2} = \sqrt{42.25 - 6.25} = \sqrt{36} = 6.00$ m (3 s.f.)
Alternatively: $h = 6.5 \sin \theta = 6.5 \times \sin 67.380...^\circ = 6.00$ m

Marks: (a) 1, (b) 2 (1 for method, 1 for answer)
Note: Use Pythagoras or trigonometry. Carry full precision for $\theta$ when using it to find $h$.

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3

(a) Using Cosine Rule:
$PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos 60^\circ$
$PR^2 = 12^2 + 9^2 - 2(12)(9)(0.5)$
$PR^2 = 144 + 81 - 108 = 117$
$PR = \sqrt{117} = 10.816... = 10.8$ cm (3 s.f.)

(b) Area $= \frac{1}{2} \times PQ \times QR \times \sin 60^\circ$
$= \frac{1}{2} \times 12 \times 9 \times \frac{\sqrt{3}}{2}$
$= 27\sqrt{3} = 46.765... = 46.8$ cm$^2$ (3 s.f.)

Marks: (a) 2 (1 for correct cosine rule, 1 for answer), (b) 1
Note: Cosine rule for side opposite known angle. Area formula $\frac{1}{2}ab\sin C$ uses included angle.

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4

(a) $\tan 35^\circ = \frac{AB}{BC} = \frac{10}{BC}$
$BC = \frac{10}{\tan 35^\circ} = \frac{10}{0.7002...} = 14.281... = 14.3$ m (3 s.f.)

(b) $\tan(\angle ADB) = \frac{AB}{DB} = \frac{10}{15} = \frac{2}{3}$
$\angle ADB = \tan^{-1}\left(\frac{2}{3}\right) = 33.690...^\circ = 33.7^\circ$ (1 d.p.)

Marks: (a) 1, (b) 2 (1 for correct ratio, 1 for answer)
Note: Angle of elevation uses tangent = opposite/adjacent. In (b), adjacent is $BD = 15$ m.

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5

(a) In right triangle $XYW$: $\sin 40^\circ = \frac{XW}{XY} = \frac{XW}{14}$
$XW = 14 \sin 40^\circ = 14 \times 0.6427... = 8.998... = 9.00$ cm (3 s.f.)

(b) Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times YZ \times XW$
$= \frac{1}{2} \times 10 \times 8.998... = 44.99... = 45.0$ cm$^2$ (3 s.f.)

Marks: (a) 1, (b) 2 (1 for method using perpendicular height, 1 for answer)
Note: $XW$ is perpendicular height to base $YZ$. Use $\sin$ since $XW$ is opposite the $40^\circ$ angle in triangle $XYW$.

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6

(a) Bearing of $Q$ from $P$ is $045^\circ$, so $\angle NPQ = 45^\circ$ where $N$ is north.
Bearing of $R$ from $Q$ is $135^\circ$, so $\angle N'QR = 135^\circ$ where $N'$ is north at $Q$.
Since north lines are parallel, $\angle PQN' = 45^\circ$ (alternate angles).
$\angle PQR = \angle N'QR - \angle PQN' = 135^\circ - 45^\circ = 90^\circ$.

(b) Triangle $PQR$ is right-angled at $Q$.
$PR^2 = PQ^2 + QR^2 = 20^2 + 15^2 = 400 + 225 = 625$
$PR = \sqrt{625} = 25.0$ km (3 s.f.)

(c) $\tan(\angle QPR) = \frac{QR}{PQ} = \frac{15}{20} = 0.75$
$\angle QPR = \tan^{-1}(0.75) = 36.869...^\circ$
Bearing of $R$ from $P$ = $045^\circ + 36.869...^\circ = 81.869...^\circ = 081.9^\circ$ (1 d.p.)

Marks: (a) 1, (b) 1, (c) 2 (1 for finding angle, 1 for bearing)
Note: Bearings are measured clockwise from north. North lines are parallel. For bearing from $P$, add $\angle QPR$ to the initial bearing $045^\circ$.

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7

(a) Triangle $AOB$ is isosceles with $OA = OB = 8$.
Drop perpendicular from $O$ to $AB$ at $M$. Then $AM = MB$ and $\angle AOM = 60^\circ$.
$\sin 60^\circ = \frac{AM}{OA} = \frac{AM}{8}$
$AM = 8 \sin 60^\circ = 8 \times \frac{\sqrt{3}}{2} = 4\sqrt{3}$
$AB = 2 \times AM = 8\sqrt{3} = 13.856... = 13.9$ cm (3 s.f.)
Alternatively using Cosine Rule: $AB^2 = 8^2 + 8^2 - 2(8)(8)\cos 120^\circ = 128 - 128(-0.5) = 192$, $AB = \sqrt{192} = 8\sqrt{3}$.

(b) Area of sector $= \frac{120}{360} \times \pi \times 8^2 = \frac{1}{3} \times 64\pi = \frac{64\pi}{3} = 67.020... = 67.0$ cm$^2$ (3 s.f.)

(c) Area of triangle $AOB = \frac{1}{2} \times 8 \times 8 \times \sin 120^\circ = 32 \times \frac{\sqrt{3}}{2} = 16\sqrt{3} = 27.712...$ cm$^2$
Area of segment = Area of sector - Area of triangle
$= 67.020... - 27.712... = 39.308... = 39.3$ cm$^2$ (3 s.f.)

Marks: (a) 1, (b) 1, (c) 2 (1 for triangle area, 1 for subtraction and answer)
Note: Segment area = sector area - triangle area. Use $\sin 120^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}$.

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8

(a) $OM = \frac{1}{2} \times BC = 5$ cm (half side of square base)
In right triangle $VOM$: $VO^2 = VM^2 - OM^2 = 13^2 - 5^2 = 169 - 25 = 144$
$VO = \sqrt{144} = 12.0$ cm (3 s.f.)

(b) Angle between face $VBC$ and base $ABCD$ is $\angle VMO$ (angle between slant height and its projection on base).
$\cos(\angle VMO) = \frac{OM}{VM} = \frac{5}{13}$
$\angle VMO = \cos^{-1}\left(\frac{5}{13}\right) = 67.380...^\circ = 67.4^\circ$ (1 d.p.)

(c) Volume $= \frac{1}{3} \times \text{base area} \times \text{height} = \frac{1}{3} \times 10^2 \times 12 = \frac{1}{3} \times 100 \times 12 = 400$ cm$^3$

Marks: (a) 1, (b) 2 (1 for identifying correct angle, 1 for answer), (c) 1
Note: Angle between plane and base = angle between line in plane perpendicular to intersection and its projection. Here $VM \perp BC$ and $OM \perp BC$, so $\angle VMO$ is the required angle.

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9

(a) Cosine Rule:
$BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos 70^\circ$
$BC^2 = 18^2 + 12^2 - 2(18)(12)\cos 70^\circ$
$BC^2 = 324 + 144 - 432(0.3420...)$
$BC^2 = 468 - 147.75... = 320.24...$
$BC = \sqrt{320.24...} = 17.895... = 17.9$ cm (3 s.f.)

(b) Sine Rule:
$\frac{\sin \angle ABC}{AC} = \frac{\sin 70^\circ}{BC}$
$\sin \angle ABC = \frac{12 \times \sin 70^\circ}{17.895...} = \frac{12 \times 0.9396...}{17.895...} = 0.6303...$
$\angle ABC = \sin^{-1}(0.6303...) = 39.06...^\circ = 39.1^\circ$ (1 d.p.)
Check: Angle is acute since $AC < AB$.

(c) Area $= \frac{1}{2} \times AB \times AC \times \sin 70^\circ$
$= \frac{1}{2} \times 18 \times 12 \times 0.9396... = 101.48... = 101$ cm$^2$ (3 s.f.)

Marks: (a) 2 (1 for cosine rule, 1 for answer), (b) 1, (c) 1
Note: Use cosine rule for side opposite known angle. Sine rule for unknown angle. Area uses included angle.

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10

(a) From triangle $ABC$: $\tan 28^\circ = \frac{h}{x + 50}$ $\Rightarrow$ $h = (x + 50)\tan 28^\circ$
From triangle $DBC$: $\tan 42^\circ = \frac{h}{x}$ $\Rightarrow$ $h = x \tan 42^\circ$

(b) Equate: $(x + 50)\tan 28^\circ = x \tan 42^\circ$
$x \tan 28^\circ + 50 \tan 28^\circ = x \tan 42^\circ$
$50 \tan 28^\circ = x(\tan 42^\circ - \tan 28^\circ)$
$x = \frac{50 \tan 28^\circ}{\tan 42^\circ - \tan 28^\circ} = \frac{50 \times 0.5317...}{0.9004... - 0.5317...} = \frac{26.585...}{0.3687...} = 72.10...$ m
$h = x \tan 42^\circ = 72.10... \times 0.9004... = 64.92... = 64.9$ m (3 s.f.)

Marks: (a) 2 (1 for each correct equation), (b) 2 (1 for solving for $x$ or $h$, 1 for final answer)
Note: Two right triangles share height $h$. Set up two tangent equations and solve simultaneously. Common error: using $x$ for $AB$ instead of $DB$.

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Section B [30 marks]

11

(a) $OM \perp AB$ (radius to midpoint of chord), so $AM = \frac{16}{2} = 8$ cm.
In right triangle $OAM$: $OM^2 = OA^2 - AM^2 = 10^2 - 8^2 = 100 - 64 = 36$
$OM = \sqrt{36} = 6$ cm

(b) $OC = 10$ cm (radius), $OM = 6$ cm
$MC = OC - OM = 10 - 6 = 4$ cm

(c) $\angle AOM = \cos^{-1}\left(\frac{OM}{OA}\right) = \cos^{-1}\left(\frac{6}{10}\right) = 53.130...^\circ$
$\angle AOB = 2 \times 53.130...^\circ = 106.260...^\circ$
Area of sector $AOB = \frac{106.260...}{360} \times \pi \times 10^2 = 92.729...$ cm$^2$
Area of triangle $AOB = \frac{1}{2} \times 10 \times 10 \times \sin 106.260...^\circ = 50 \times 0.96 = 48$ cm$^2$
Area of segment $= 92.729... - 48 = 44.729... = 44.7$ cm$^2$ (3 s.f.)

Marks: (a) 1, (b) 1, (c) 3 (1 for angle, 1 for sector area, 1 for triangle area and subtraction)
Note: Perpendicular from centre bisects chord. Central angle found via cosine. Segment = sector - triangle.

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12

(a) $\angle ADE = 180^\circ - 110^\circ = 70^\circ$ (angles on straight line)
In right triangle $ADE$: $\sin 70^\circ = \frac{AE}{AD} = \frac{AE}{10}$
$AE = 10 \sin 70^\circ = 10 \times 0.9396... = 9.396... = 9.40$ cm (3 s.f.)

(b) $\cos 70^\circ = \frac{DE}{AD} = \frac{DE}{10}$
$DE = 10 \cos 70^\circ = 10 \times 0.3420... = 3.420... = 3.42$ cm (3 s.f.)

(c) $EC = DC - DE = 12 - 3.420... = 8.579...$ cm
Area of trapezium $= \frac{1}{2}(AB + DC) \times AE = \frac{1}{2}(20 + 12) \times 9.396...$
$= 16 \times 9.396... = 150.34... = 150$ cm$^2$ (3 s.f.)

Marks: (a) 1, (b) 1, (c) 3 (1 for finding $EC$ or using trapezium formula, 1 for correct substitution, 1 for answer)
Note: $\angle ADC = 110^\circ$ is obtuse, so the interior angle at $D$ for the right triangle is $70^\circ$. Height $AE$ is perpendicular to both parallel sides.

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13

(a) In right triangle $PAQ$: $\tan 30^\circ = \frac{h}{AQ}$ $\Rightarrow$ $AQ = \frac{h}{\tan 30^\circ} = h\sqrt{3}$
In right triangle $PBQ$: $\tan 25^\circ = \frac{h}{BQ}$ $\Rightarrow$ $BQ = \frac{h}{\tan 25^\circ}$

(b) Triangle $AQB$ is right-angled at $Q$ (south and east are perpendicular).
$AB^2 = AQ^2 + BQ^2$
$100^2 = (h\sqrt{3})^2 + \left(\frac{h}{\tan 25^\circ}\right)^2$
$10000 = 3h^2 + \frac{h^2}{\tan^2 25^\circ}$
$10000 = h^2\left(3 + \frac{1}{0.4663

<stage3_exam_answers_md>

TuitionGoWhere Practice Paper - Elementary Mathematics Secondary 3 (SA2 Version 2) - Answer Key

Total Marks: 60

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Section A [30 marks]

1

(a) $\tan x = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AB} = \frac{8}{15}$

(b) $x = \tan^{-1}\left(\frac{8}{15}\right) = 28.072...^\circ = 28.1^\circ$ (1 d.p.)

Marks: (a) 1, (b) 1
Note: Identify opposite and adjacent correctly relative to angle $x$. Angle $x$ is at $A$, so opposite is $BC = 8$, adjacent is $AB = 15$.

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2

(a) $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{2.5}{6.5} = \frac{5}{13}$
$\theta = \cos^{-1}\left(\frac{5}{13}\right) = 67.380...^\circ = 67.4^\circ$ (1 d.p.)

(b) Height $h = \sqrt{6.5^2 - 2.5^2} = \sqrt{42.25 - 6.25} = \sqrt{36} = 6.00$ m (3 s.f.)
Alternatively: $h = 6.5 \sin \theta = 6.5 \times \sin 67.380...^\circ = 6.00$ m

Marks: (a) 1, (b) 2 (1 for method, 1 for answer)
Note: Use Pythagoras or trigonometry. Carry full precision for $\theta$ when using it to find $h$.

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3

(a) Using Cosine Rule:
$PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos 60^\circ$
$PR^2 = 12^2 + 9^2 - 2(12)(9)(0.5)$
$PR^2 = 144 + 81 - 108 = 117$
$PR = \sqrt{117} = 10.816... = 10.8$ cm (3 s.f.)

(b) Area $= \frac{1}{2} \times PQ \times QR \times \sin 60^\circ$
$= \frac{1}{2} \times 12 \times 9 \times \frac{\sqrt{3}}{2}$
$= 27\sqrt{3} = 46.765... = 46.8$ cm$^2$ (3 s.f.)

Marks: (a) 2 (1 for correct cosine rule, 1 for answer), (b) 1
Note: Cosine rule for side opposite known angle. Area formula $\frac{1}{2}ab\sin C$ uses included angle.

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4

(a) $\tan 35^\circ = \frac{AB}{BC} = \frac{10}{BC}$
$BC = \frac{10}{\tan 35^\circ} = \frac{10}{0.7002...} = 14.281... = 14.3$ m (3 s.f.)

(b) $\tan(\angle ADB) = \frac{AB}{DB} = \frac{10}{15} = \frac{2}{3}$
$\angle ADB = \tan^{-1}\left(\frac{2}{3}\right) = 33.690...^\circ = 33.7^\circ$ (1 d.p.)

Marks: (a) 1, (b) 2 (1 for correct ratio, 1 for answer)

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5

(a) In right triangle $XYW$, $\sin 40^\circ = \frac{XW}{XY} = \frac{XW}{14}$
$XW = 14 \sin 40^\circ = 14 \times 0.6427... = 8.998... = 9.00$ cm (3 s.f.)

(b) Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times YZ \times XW$
$= \frac{1}{2} \times 10 \times 8.998... = 44.99... = 45.0$ cm$^2$ (3 s.f.)
Alternatively: Area $= \frac{1}{2} \times XY \times YZ \times \sin 40^\circ = \frac{1}{2} \times 14 \times 10 \times \sin 40^\circ = 45.0$ cm$^2$

Marks: (a) 1, (b) 2 (1 for method, 1 for answer)

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6

(a) Bearing of $Q$ from $P$ is $045^\circ$, so North line at $P$ to $PQ$ is $45^\circ$.
Bearing of $R$ from $Q$ is $135^\circ$, so North line at $Q$ to $QR$ is $135^\circ$.
Since North lines are parallel, interior angles sum to $180^\circ$.
Angle between $PQ$ and South at $Q$ = $45^\circ$ (alternate angles).
$\angle PQR = 135^\circ - 45^\circ = 90^\circ$. (Shown)

(b) Triangle $PQR$ is right-angled at $Q$.
$PR^2 = PQ^2 + QR^2 = 20^2 + 15^2 = 400 + 225 = 625$
$PR = \sqrt{625} = 25.0$ km (3 s.f.)

(c) $\tan(\angle QPR) = \frac{QR}{PQ} = \frac{15}{20} = 0.75$
$\angle QPR = \tan^{-1}(0.75) = 36.869...^\circ$
Bearing of $R$ from $P$ = $045^\circ + 36.869...^\circ = 081.869...^\circ = 081.9^\circ$ (1 d.p.)

Marks: (a) 1, (b) 2 (1 for Pythagoras, 1 for answer), (c) 1

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7

(a) Chord length $AB = 2r \sin\left(\frac{\theta}{2}\right) = 2 \times 8 \times \sin 60^\circ = 16 \times \frac{\sqrt{3}}{2} = 8\sqrt{3} = 13.856... = 13.9$ cm (3 s.f.)
Alternatively: Cosine rule in $\triangle AOB$: $AB^2 = 8^2 + 8^2 - 2(8)(8)\cos 120^\circ = 128 - 128(-0.5) = 192$, $AB = \sqrt{192} = 13.9$ cm.

(b) Area of sector $= \frac{120}{360} \times \pi r^2 = \frac{1}{3} \times \pi \times 64 = \frac{64\pi}{3} = 67.020... = 67.0$ cm$^2$ (3 s.f.)

(c) Area of $\triangle AOB = \frac{1}{2} r^2 \sin 120^\circ = \frac{1}{2} \times 64 \times \frac{\sqrt{3}}{2} = 16\sqrt{3} = 27.712...$ cm$^2$
Area of segment = Area of sector $-$ Area of triangle
$= 67.020... - 27.712... = 39.308... = 39.3$ cm$^2$ (3 s.f.)

Marks: (a) 1, (b) 1, (c) 2 (1 for triangle area, 1 for segment area)

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8

(a) $OM = \frac{10}{2} = 5$ cm (half side of square base)
In right $\triangle VOM$, $VO^2 = VM^2 - OM^2 = 13^2 - 5^2 = 169 - 25 = 144$
$VO = \sqrt{144} = 12.0$ cm (3 s.f.)

(b) Angle between face $VBC$ and base $ABCD$ is $\angle VMO$ (angle between slant height and its projection).
$\tan(\angle VMO) = \frac{VO}{OM} = \frac{12}{5} = 2.4$
$\angle VMO = \tan^{-1}(2.4) = 67.380...^\circ = 67.4^\circ$ (1 d.p.)

(c) Volume $= \frac{1}{3} \times \text{base area} \times \text{height} = \frac{1}{3} \times 10^2 \times 12 = \frac{1}{3} \times 100 \times 12 = 400$ cm$^3$ (exact, 3 s.f. = 400)

Marks: (a) 2 (1 for $OM=5$, 1 for $VO$), (b) 1, (c) 1

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9

(a) Cosine Rule:
$BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos 70^\circ$
$BC^2 = 18^2 + 12^2 - 2(18)(12)\cos 70^\circ$
$BC^2 = 324 + 144 - 432 \times 0.3420... = 468 - 147.75... = 320.24...$
$BC = \sqrt{320.24...} = 17.895... = 17.9$ cm (3 s.f.)

(b) Sine Rule:
$\frac{\sin \angle ABC}{AC} = \frac{\sin 70^\circ}{BC}$
$\sin \angle ABC = \frac{12 \times \sin 70^\circ}{17.895...} = \frac{12 \times 0.9396...}{17.895...} = 0.6303...$
$\angle ABC = \sin^{-1}(0.6303...) = 39.07...^\circ = 39.1^\circ$ (1 d.p.)
(Note: Angle is acute since $AC < AB$)

(c) Area $= \frac{1}{2} \times AB \times AC \times \sin 70^\circ = \frac{1}{2} \times 18 \times 12 \times 0.9396... = 101.48... = 101$ cm$^2$ (3 s.f.)

Marks: (a) 2, (b) 1, (c) 1

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10

(a) From $\triangle ABC$: $\tan 28^\circ = \frac{h}{x + 50} \implies h = (x + 50)\tan 28^\circ$
From $\triangle DBC$: $\tan 42^\circ = \frac{h}{x} \implies h = x \tan 42^\circ$

(b) Equate: $x \tan 42^\circ = (x + 50)\tan 28^\circ$
$x(\tan 42^\circ - \tan 28^\circ) = 50 \tan 28^\circ$
$x = \frac{50 \tan 28^\circ}{\tan 42^\circ - \tan 28^\circ} = \frac{50 \times 0.5317...}{0.9004... - 0.5317...} = \frac{26.585...}{0.3687...} = 72.10...$ m
$h = x \tan 42^\circ = 72.10... \times 0.9004... = 64.92... = 64.9$ m (3 s.f.)

Marks: (a) 2 (1 for each equation), (b) 2 (1 for solving $x$, 1 for $h$)

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Section B [30 marks]

11

(a) $OM \perp AB$, so $AM = \frac{16}{2} = 8$ cm.
In right $\triangle OAM$, $OM^2 = OA^2 - AM^2 = 10^2 - 8^2 = 100 - 64 = 36$
$OM = 6$ cm

(b) $OC = 10$ cm (radius), $MC = OC - OM = 10 - 6 = 4$ cm

(c) Area of sector $AOB$: $\angle AOB = 2 \sin^{-1}\left(\frac{AM}{OA}\right) = 2 \sin^{-1}(0.8) = 106.26...^\circ$
Sector area $= \frac{106.26...}{360} \times \pi \times 10^2 = 92.729...$ cm$^2$
Area of $\triangle AOB = \frac{1}{2} \times AB \times OM = \frac{1}{2} \times 16 \times 6 = 48$ cm$^2$
Segment area $= 92.729... - 48 = 44.729... = 44.7$ cm$^2$ (3 s.f.)

Marks: (a) 1, (b) 1, (c) 3 (1 for angle/sector area, 1 for triangle area, 1 for segment)

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12

(a) $AE = AD \sin \angle ADC = 10 \sin 110^\circ = 10 \sin 70^\circ = 10 \times 0.9396... = 9.396... = 9.40$ cm (3 s.f.)

(b) $DE = AD \cos \angle ADC = 10 \cos 110^\circ = -10 \cos 70^\circ = -10 \times 0.3420... = -3.420...$
Length $DE = 3.42$ cm (3 s.f.) (The negative sign indicates $E$ lies on extension of $DC$ past $D$)

(c) Since $AB \parallel DC$, height of trapezium $= AE = 9.396...$ cm
Area $= \frac{1}{2}(AB + DC) \times h = \frac{1}{2}(20 + 12) \times 9.396... = 16 \times 9.396... = 150.34... = 150$ cm$^2$ (3 s.f.)

Marks: (a) 1, (b) 1, (c) 3 (1 for height, 1 for formula, 1 for answer)

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13

(a) In $\triangle PAQ$, $\tan 30^\circ = \frac{h}{AQ} \implies AQ = \frac{h}{\tan 30^\circ} = h\sqrt{3}$
In $\triangle PBQ$, $\tan 25^\circ = \frac{h}{BQ} \implies BQ = \frac{h}{\tan 25^\circ}$

(b) $\triangle AQB$ is right-angled at $Q$ (South and East are perpendicular).
$AB^2 = AQ^2 + BQ^2$
$100^2 = (h\sqrt{3})^2 + \left(\frac{h}{\tan 25^\circ}\right)^2$
$10000 = 3h^2 + \frac{h^2}{\tan^2 25^\circ}$
$10000 = h^2\left(3 + \frac{1}{0.4663...^2}\right) = h^2(3 + 4.599...) = 7.599... h^2$
$h^2 = \frac{10000}{7.599...} = 1315.8...$
$h = \sqrt{1315.8...} = 36.27... = 36.3$ m (3 s.f.)

(c) Midpoint $M$ of $AB$: $MQ = \frac{1}{2}AB = 50$ m (midpoint of hypotenuse in right triangle).
$\tan(\angle PMQ) = \frac{h}{MQ} = \frac{36.27...}{50} = 0.7255...$
$\angle PMQ = \tan^{-1}(0.7255...) = 35.97...^\circ = 36.0^\circ$ (1 d.p.)

Marks: (a) 2, (b) 3 (1 for Pythagoras, 1 for equation, 1 for answer), (c) 1

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14

(a) Slant height $l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ cm

(b) Total Surface Area = Curved surface of cone + Curved surface of hemisphere
$= \pi r l + 2\pi r^2 = \pi(6)(10) + 2\pi(6^2) = 60\pi + 72\pi = 132\pi = 414.69... = 415$ cm$^2$ (3 s.f.)
(Note: Base of cone and flat face of hemisphere are internal, not counted)

(c) Volume = Volume of cone + Volume of hemisphere
$= \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi(36)(8) + \frac{2}{3}\pi(216) = 96\pi + 144\pi = 240\pi = 753.98... = 754$ cm$^3$ (3 s.f.)

Marks: (a) 1, (b) 2 (1 for cone CSA, 1 for hemisphere CSA + total), (c) 2 (1 for cone vol, 1 for hemisphere vol + total)

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15

(a) $\angle AOC = \angle AOB + \angle BOC = 80^\circ + 100^\circ = 180^\circ$
(So $AC$ is a diameter)

(b) $\angle ABC$ is angle at circumference subtended by arc $AC$ (major arc).
Reflex $\angle AOC = 360^\circ - 180^\circ = 180^\circ$.
$\angle ABC = \frac{1}{2} \times \text{reflex } \angle AOC = \frac{1}{2} \times 180^\circ = 90^\circ$
(Alternatively: Angle in semicircle = $90^\circ$)

(c) $AD = DC \implies$ arc $AD$ = arc $DC \implies \angle AOD = \angle DOC = \frac{180^\circ}{2} = 90^\circ$.
$\angle ADC$ is angle at circumference subtended by arc $ABC$ (major arc $AC$ via $B$).
Reflex $\angle AOC$ (via $B$) = $360^\circ - 180^\circ = 180^\circ$? Wait.
Arc $ABC$ corresponds to central angle $\angle AOB + \angle BOC = 180^\circ$.
$\angle ADC = \frac{1}{2} \times (\text{angle subtended by arc } ABC \text{ at centre}) = \frac{1}{2} \times 180^\circ = 90^\circ$.
Alternatively: Cyclic quadrilateral $ABCD$: $\angle ABC + \angle ADC = 180^\circ \implies 90^\circ + \angle ADC = 180^\circ \implies \angle ADC = 90^\circ$.

(d) $AC$ is diameter $= 2 \times 12 = 24$ cm.
(Or chord length: $2r \sin(\frac{180^\circ}{2}) = 24 \sin 90^\circ = 24$ cm)

Marks: (a) 1, (b) 1, (c) 1, (d) 2 (1 for diameter recognition, 1 for answer)

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16

(a) Bearing $A \to B = 060^\circ$. North at $B$ parallel to North at $A$.
Angle between $AB$ and North at $B$ (back-bearing) = $060^\circ$ (alternate angles).
Bearing $B \to C = 150^\circ$.
$\angle ABC = 150^\circ - 60^\circ = 90^\circ$. (Shown)

(b) Right triangle $ABC$, right angle at $B$.
$CA^2 = AB^2 + BC^2 = 500^2 + 400^2 = 250000 + 160000 = 410000$
$CA = \sqrt{410000} = 640.31... = 640$ m (3 s.f.)

(c) $\tan(\angle BAC) = \frac{BC}{AB} = \frac{400}{500} = 0.8$
$\angle BAC = \tan^{-1}(0.8) = 38.659...^\circ$
Bearing of $A$ from $C$: Need angle clockwise from North at $C$ to $CA$.
At $C$, North line parallel to North at $A$.
Angle between $CA$ and South at $C$ = $\angle BAC = 38.659...^\circ$ (alternate angles).
Bearing = $180^\circ + 38.659...^\circ = 218.659...^\circ = 218.7^\circ$ (1 d.p.)

(d) Total distance $= AB + BC + CA = 500 + 400 + 640.31... = 1540.31... = 1540$ m (3 s.f.)

Marks: (a) 1, (b) 2, (c) 2 (1 for $\angle BAC$, 1 for bearing), (d) 1

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END OF ANSWER KEY