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

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TuitionGoWhere Practice Paper — Elementary Mathematics Secondary 3

School: TuitionGoWhere Secondary School (AI)

Subject: Elementary Mathematics Level: Secondary 3 Paper: SA2 Practice — Version 5 of 5 Duration: 60 minutes Total Marks: 50

Name: ___________________________ Class: ___________________________ Date: ___________________________

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Instructions

• Answer ALL questions in the spaces provided.
• Show all working clearly. Marks are awarded for correct method even if the final answer is incorrect.
• The use of calculators is allowed unless stated otherwise.
• Unless otherwise stated, give non-exact answers correct to 1 decimal place.
• Do not use correction fluid or tape.

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Section A — Short Answer Questions [20 marks]

Answer ALL questions. Each question carries 2 marks unless otherwise stated.

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1. In right-angled triangle $PQR$, $\angle Q = 90^\circ$, $PQ = 12$ cm and $QR = 5$ cm. Calculate $\angle PRQ$, giving your answer correct to 1 decimal place.

 

 

 

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2. In right-angled triangle $ABC$, $\angle B = 90^\circ$, $AB = 7$ cm and $AC = 25$ cm. Calculate the length of $BC$.

 

 

 

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3. A ladder 8 m long leans against a vertical wall. The foot of the ladder is 3.5 m from the base of the wall. Calculate the angle the ladder makes with the ground, giving your answer correct to 1 decimal place.

 

 

 

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4. In $\triangle XYZ$, $\angle X = 90^\circ$, $XY = 9$ cm and $\tan \angle YZX = \frac{3}{4}$. Calculate the length of $XZ$.

 

 

 

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5. From a point $P$ on horizontal ground, the angle of elevation to the top of a building is $38^\circ$. The distance from $P$ to the base of the building is 65 m. Calculate the height of the building, giving your answer correct to 1 decimal place.

 

 

 

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6. In right-angled triangle $DEF$, $\angle E = 90^\circ$, $DE = 15$ cm and $\sin \angle DFE = \frac{5}{13}$. Calculate the length of $EF$.

 

 

 

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7. A vertical pole $AB$ stands on horizontal ground. From a point $C$ on the ground, 20 m from the base of the pole, the angle of elevation to the top of the pole is $52^\circ$. Calculate the height of the pole, giving your answer correct to 1 decimal place.

 

 

 

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8. In $\triangle LMN$, $\angle M = 90^\circ$, $LM = 8$ cm and $LN = 17$ cm. Calculate $\angle LNM$, giving your answer correct to 1 decimal place.

 

 

 

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9. A ship sails 12 km due east from port $A$ to point $B$, then sails 9 km due north to point $C$. Calculate the bearing of $C$ from $A$, giving your answer correct to the nearest degree.

 

 

 

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10. In right-angled triangle $WXY$, $\angle W = 90^\circ$, $\cos \angle WYX = \frac{7}{25}$ and $WY = 21$ cm. Calculate the length of $WX$.

 

 

 

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Section B — Structured Questions [20 marks]

Answer ALL questions. Show all working clearly.

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11. The diagram shows right-angled triangle $ABC$ with $\angle B = 90^\circ$, $AB = 16$ cm and $BC = 30$ cm.

(a) Calculate the length of $AC$. [2]

 

 

 

(b) Calculate $\angle ACB$, giving your answer correct to 1 decimal place. [2]

 

 

 

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12. A flagpole stands on horizontal ground. From a point $P$ on the ground, the angle of elevation to the top of the flagpole is $40^\circ$. From a point $Q$, which is 15 m further away from the flagpole along the same straight line, the angle of elevation is $25^\circ$.

(a) Using the information, write down an expression for the height $h$ of the flagpole in terms of the distance from $P$ to the base of the flagpole. [1]

 

 

(b) Hence, calculate the height of the flagpole, giving your answer correct to 1 decimal place. [3]

 

 

 

 

 

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13. In $\triangle PQR$, $\angle Q = 90^\circ$, $PQ = (3x)$ cm, $QR = (4x)$ cm and $PR = 30$ cm.

(a) Form an equation in $x$ and solve it. [2]

 

 

 

(b) Hence, calculate $\angle PRQ$, giving your answer correct to 1 decimal place. [2]

 

 

 

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14. A vertical cliff is 80 m high. From the top of the cliff, the angle of depression of a boat at sea is $28^\circ$.

(a) Explain why the angle of depression from the top of the cliff equals the angle of elevation from the boat. [1]

 

 

(b) Calculate the distance of the boat from the base of the cliff, giving your answer correct to 1 decimal place. [3]

 

 

 

 

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15. The bearing of $B$ from $A$ is $065^\circ$. The bearing of $C$ from $B$ is $155^\circ$. $AB = 24$ km and $BC = 18$ km.

(a) Find $\angle ABC$. [2]

 

 

(b) Calculate the distance $AC$, giving your answer correct to 1 decimal place. [2]

 

 

 

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Section C — Application Problem [10 marks]

Answer the question. Show all working clearly.

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16. A communications tower $TS$ stands on horizontal ground. From a point $A$ due south of the tower, the angle of elevation to the top $T$ of the tower is $50^\circ$. From a point $B$ due west of the tower, the angle of elevation to the top of the tower is $35^\circ$. The distance $AB$ is 120 m.

(a) Let the height of the tower be $h$ metres. Write expressions for $AS$ and $BS$ in terms of $h$. [2]

 

 

 

(b) Using your answers in (a), show that $h^2\left(\frac{1}{\tan^2 50^\circ} + \frac{1}{\tan^2 35^\circ}\right) = 120^2$. [2]

 

 

 

 

(c) Hence, calculate the height of the tower, giving your answer correct to 1 decimal place. [2]

 

 

 

(d) Calculate the bearing of $B$ from $A$. [2]

 

 

 

(e) A bird sits at the top of the tower. Calculate the angle of depression of point $A$ from the top of the tower. [2]

 

 

 

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17. In $\triangle ABC$, $\angle C = 90^\circ$, $AC = 12$ cm and $BC = 5$ cm. Point $D$ lies on $AB$ such that $CD$ is perpendicular to $AB$.

(a) Calculate the length of $AB$. [1]

 

 

(b) Using the area of $\triangle ABC$, calculate the length of $CD$. [2]

 

 

 

(c) Calculate $\angle ACD$, giving your answer correct to 1 decimal place. [2]

 

 

 

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18. From the top of a building 60 m tall, the angles of depression of two cars on a straight road at ground level are $18^\circ$ and $32^\circ$. Both cars are on the same side of the building.

(a) Calculate the distance of each car from the base of the building. [2]

 

 

 

(b) Calculate the distance between the two cars, giving your answer correct to 1 decimal place. [2]

 

 

 

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19. A triangular plot of land $PQR$ has $\angle P = 90^\circ$, $PQ = 45$ m and $\angle PRQ = 22^\circ$.

(a) Calculate the length of $PR$, giving your answer correct to 1 decimal place. [2]

 

 

 

(b) Calculate the area of the plot, giving your answer correct to 1 decimal place. [2]

 

 

 

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20. A plane flies from town $X$ to town $Y$, a distance of 350 km, on a bearing of $048^\circ$. It then flies from town $Y$ to town $Z$ on a bearing of $138^\circ$. The distance $YZ$ is 280 km.

(a) Calculate the angle $\angle XYZ$. [2]

 

 

 

(b) Calculate the distance $XZ$, giving your answer correct to 1 decimal place. [2]

 

 

 

(c) Calculate the bearing of $X$ from $Z$, giving your answer to the nearest degree. [2]

 

 

 

 

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— End of Paper —

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TuitionGoWhere Practice Paper — Elementary Mathematics Secondary 3

Answer Key — Version 5 of 5

Paper: SA2 Practice | Topic: Geometry & Trigonometry | Total Marks: 50

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Section A — Short Answer Questions

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1. [2 marks]

In $\triangle PQR$, $\angle Q = 90^\circ$, $PQ = 12$ cm, $QR = 5$ cm.

$\tan(\angle PRQ) = \frac{PQ}{QR} = \frac{12}{5} = 2.4$

$\angle PRQ = \tan^{-1}(2.4) = 67.380...^\circ$

$\boxed{\angle PRQ = 67.4^\circ}$

Marking: M1 for correct trig ratio setup; A1 for answer to 1 d.p.

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2. [2 marks]

In $\triangle ABC$, $\angle B = 90^\circ$, $AB = 7$ cm, $AC = 25$ cm.

By Pythagoras' theorem: $BC^2 = AC^2 - AB^2 = 25^2 - 7^2 = 625 - 49 = 576$

$BC = \sqrt{576} = 24$

$\boxed{BC = 24 \text{ cm}}$

Marking: M1 for correct Pythagoras setup; A1 for 24 cm.

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3. [2 marks]

Let $\theta$ be the angle the ladder makes with the ground.

$\cos\theta = \frac{3.5}{8} = 0.4375$

$\theta = \cos^{-1}(0.4375) = 64.055...^\circ$

$\boxed{\theta = 64.1^\circ}$

Marking: M1 for correct trig ratio; A1 for answer to 1 d.p.

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4. [2 marks]

In $\triangle XYZ$, $\angle X = 90^\circ$, $XY = 9$ cm.

$\tan(\angle YZX) = \frac{XY}{XZ} = \frac{3}{4}$

$\frac{9}{XZ} = \frac{3}{4}$

$XZ = \frac{9 \times 4}{3} = 12$

$\boxed{XZ = 12 \text{ cm}}$

Marking: M1 for correct tan ratio setup; A1 for 12 cm.

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5. [2 marks]

Let $h$ be the height of the building.

$\tan 38^\circ = \frac{h}{65}$

$h = 65 \times \tan 38^\circ = 65 \times 0.781285... = 50.783...$

$\boxed{h = 50.8 \text{ m}}$

Marking: M1 for $\tan 38^\circ = h/65$; A1 for 50.8 m.

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6. [2 marks]

In $\triangle DEF$, $\angle E = 90^\circ$, $DE = 15$ cm.

$\sin(\angle DFE) = \frac{DE}{DF} = \frac{5}{13}$

$\frac{15}{DF} = \frac{5}{13}$

$DF = \frac{15 \times 13}{5} = 39$ cm

By Pythagoras: $EF^2 = DF^2 - DE^2 = 39^2 - 15^2 = 1521 - 225 = 1296$

$EF = \sqrt{1296} = 36$

$\boxed{EF = 36 \text{ cm}}$

Marking: M1 for finding DF using sin ratio; A1 for EF = 36 cm.

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7. [2 marks]

Let $h$ be the height of the pole.

$\tan 52^\circ = \frac{h}{20}$

$h = 20 \times \tan 52^\circ = 20 \times 1.27994... = 25.598...$

$\boxed{h = 25.6 \text{ m}}$

Marking: M1 for $\tan 52^\circ = h/20$; A1 for 25.6 m.

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8. [2 marks]

In $\triangle LMN$, $\angle M = 90^\circ$, $LM = 8$ cm, $LN = 17$ cm.

By Pythagoras: $MN^2 = LN^2 - LM^2 = 17^2 - 8^2 = 289 - 64 = 225$

$MN = \sqrt{225} = 15$ cm

$\tan(\angle LNM) = \frac{LM}{MN} = \frac{8}{15}$

$\angle LNM = \tan^{-1}\left(\frac{8}{15}\right) = 28.072...^\circ$

$\boxed{\angle LNM = 28.1^\circ}$

Marking: M1 for finding MN = 15 and setting up tan ratio; A1 for 28.1°.

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9. [2 marks]

The ship sails 12 km east then 9 km north, forming a right angle at $B$.

$\tan(\angle CAB) = \frac{9}{12} = 0.75$

$\angle CAB = \tan^{-1}(0.75) = 36.869...^\circ$

Bearing of $C$ from $A = 90^\circ - 36.869... = 53.130...^\circ$

$\boxed{\text{Bearing} = 053^\circ}$

Marking: M1 for correct tan calculation; A1 for bearing 053° (nearest degree).

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10. [2 marks]

In $\triangle WXY$, $\angle W = 90^\circ$, $\cos(\angle WYX) = \frac{7}{25}$, $WY = 21$ cm.

$\cos(\angle WYX) = \frac{WY}{WY} = \frac{7}{25}$ — wait, adjacent to $\angle WYX$ is $WY$ and hypotenuse is $XY$.

$\cos(\angle WYX) = \frac{WY}{XY} = \frac{7}{25}$

$\frac{21}{XY} = \frac{7}{25}$

$XY = \frac{21 \times 25}{7} = 75$ cm

By Pythagoras: $WX^2 = XY^2 - WY^2 = 75^2 - 21^2 = 5625 - 441 = 5184$

$WX = \sqrt{5184} = 72$

$\boxed{WX = 72 \text{ cm}}$

Marking: M1 for finding XY = 75 using cos ratio; A1 for WX = 72 cm.

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Section B — Structured Questions

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11. [4 marks total]

(a) [2 marks]

In $\triangle ABC$, $\angle B = 90^\circ$, $AB = 16$ cm, $BC = 30$ cm.

$AC^2 = AB^2 + BC^2 = 16^2 + 30^2 = 256 + 900 = 1156$

$AC = \sqrt{1156} = 34$

$\boxed{AC = 34 \text{ cm}}$

Marking: M1 for Pythagoras setup; A1 for 34 cm.

(b) [2 marks]

$\tan(\angle ACB) = \frac{AB}{BC} = \frac{16}{30} = \frac{8}{15}$

$\angle ACB = \tan^{-1}\left(\frac{8}{15}\right) = 28.072...^\circ$

$\boxed{\angle ACB = 28.1^\circ}$

Marking: M1 for correct tan ratio; A1 for 28.1°.

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12. [4 marks total]

(a) [1 mark]

Let the distance from $P$ to the base of the flagpole be $x$ metres.

$\tan 40^\circ = \frac{h}{x}$

$\boxed{h = x \tan 40^\circ}$

Mark for correct expression.

(b) [3 marks]

From point $Q$, distance from base $= x + 15$:

$\tan 25^\circ = \frac{h}{x + 15}$

$h = (x + 15)\tan 25^\circ$

Equating: $x \tan 40^\circ = (x + 15)\tan 25^\circ$

$x \tan 40^\circ = x \tan 25^\circ + 15 \tan 25^\circ$

$x(\tan 40^\circ - \tan 25^\circ) = 15 \tan 25^\circ$

$x(0.8391 - 0.4663) = 15 \times 0.4663$

$x(0.3728) = 6.9945$

$x = \frac{6.9945}{0.3728} = 18.763...$

$h = 18.763 \times \tan 40^\circ = 18.763 \times 0.8391 = 15.743...$

$\boxed{h = 15.7 \text{ m}}$

Marking: M1 for setting up second equation; M1 for solving for x; A1 for h = 15.7 m.

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13. [4 marks total]

(a) [2 marks]

By Pythagoras' theorem:

$(3x)^2 + (4x)^2 = 30^2$

$9x^2 + 16x^2 = 900$

$25x^2 = 900$

$x^2 = 36$

$x = 6$ (reject negative)

$\boxed{x = 6}$

Marking: M1 for correct equation; A1 for x = 6.

(b) [2 marks]

$PQ = 3(6) = 18$ cm, $QR = 4(6) = 24$ cm.

$\tan(\angle PRQ) = \frac{PQ}{QR} = \frac{18}{24} = \frac{3}{4}$

$\angle PRQ = \tan^{-1}(0.75) = 36.869...^\circ$

$\boxed{\angle PRQ = 36.9^\circ}$

Marking: M1 for correct tan ratio; A1 for 36.9°.

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14. [4 marks total]

(a) [1 mark]

The angle of depression from the top of the cliff equals the angle of elevation from the boat because the line of sight and the horizontal lines at the top of the cliff and at the boat are parallel, and the transversal creates equal alternate angles.

$\boxed{\text{Alternate angles between parallel horizontals are equal.}}$

Mark for mentioning alternate angles or parallel lines.

(b) [3 marks]

Let $d$ be the distance of the boat from the base of the cliff.

$\tan 28^\circ = \frac{80}{d}$

$d = \frac{80}{\tan 28^\circ} = \frac{80}{0.5317} = 150.458...$

$\boxed{d = 150.5 \text{ m}}$

Marking: M1 for correct tan setup; M1 for rearranging; A1 for 150.5 m.

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15. [4 marks total]

(a) [2 marks]

Bearing of $B$ from $A$ is $065^\circ$, so the direction $AN$ (north at $A$) to $AB$ is $65^\circ$. Bearing of $C$ from $B$ is $155^\circ$.

At point $B$, the back-bearing of $A$ from $B$ is $180^\circ + 65^\circ = 245^\circ$ (or equivalently, the angle between $BA$ (towards $A$) and north at $B$ is $65^\circ$ measured the other way).

The angle between $BA$ and $BC$:

At $B$, the angle from north to $BA$ (back-bearing direction towards $A$) is $65^\circ + 180^\circ = 245^\circ$. The bearing of $C$ from $B$ is $155^\circ$.

$\angle ABC = 245^\circ - 155^\circ = 90^\circ$

Alternatively: The bearing of $A$ from $B$ is $065^\circ + 180^\circ = 245^\circ$. The angle between direction $BA$ (bearing $245^\circ$) and direction $BC$ (bearing $155^\circ$) is $245^\circ - 155^\circ = 90^\circ$.

$\boxed{\angle ABC = 90^\circ}$

Marking: M1 for finding back-bearing; A1 for 90°.

(b) [2 marks]

Since $\angle ABC = 90^\circ$, $\triangle ABC$ is right-angled at $B$.

$AC^2 = AB^2 + BC^2 = 24^2 + 18^2 = 576 + 324 = 900$

$AC = \sqrt{900} = 30$

$\boxed{AC = 30.0 \text{ km}}$

Marking: M1 for Pythagoras; A1 for 30.0 km.

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Section C — Application Problem

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16. [10 marks total]

(a) [2 marks]

From point $A$ (due south): $\tan 50^\circ = \frac{h}{AS}$, so $AS = \frac{h}{\tan 50^\circ} = h \cot 50^\circ$

From point $B$ (due west): $\tan 35^\circ = \frac{h}{BS}$, so $BS = \frac{h}{\tan 35^\circ} = h \cot 35^\circ$

$\boxed{AS = h \cot 50^\circ \quad \text{and} \quad BS = h \cot 35^\circ}$

Marking: 1 mark each for correct expressions.

(b) [2 marks]

Since $A$ is due south and $B$ is due west of $S$, $\angle ASB = 90^\circ$.

By Pythagoras in $\triangle ASB$:

$AS^2 + BS^2 = AB^2$

$(h \cot 50^\circ)^2 + (h \cot 35^\circ)^2 = 120^2$

$h^2 \cot^2 50^\circ + h^2 \cot^2 35^\circ = 14400$

$h^2\left(\frac{1}{\tan^2 50^\circ} + \frac{1}{\tan^2 35^\circ}\right) = 120^2$

$\boxed{h^2\left(\frac{1}{\tan^2 50^\circ} + \frac{1}{\tan^2 35^\circ}\right) = 14400}$

Marking: M1 for Pythagoras on triangle ASB; A1 for correct substitution.

(c) [2 marks]

$\frac{1}{\tan^2 50^\circ} = \frac{1}{1.1918^2} = \frac{1}{1.4203} = 0.7041$

$\frac{1}{\tan^2 35^\circ} = \frac{1}{0.7002^2} = \frac{1}{0.4903} = 2.0396$

$h^2(0.7041 + 2.0396) = 14400$

$h^2(2.7437) = 14400$

$h^2 = \frac{14400}{2.7437} = 5248.39...$

$h = \sqrt{5248.39} = 72.445...$

$\boxed{h = 72.4 \text{ m}}$

Marking: M1 for correct substitution and evaluation; A1 for 72.4 m.

(d) [2 marks]

Point $A$ is due south of $S$ and point $B$ is due west of $S$.

So from $A$, point $B$ is to the north-west. Specifically, $S$ is north of $A$, and $B$ is west of $S$.

From $A$: $S$ is due north. From $S$, $B$ is due west. So from $A$, $B$ is at bearing $270^\circ +$ (angle from north at A to line AB).

In $\triangle ASB$, $\angle SAB$: $\tan(\angle SAB) = \frac{BS}{AS}$

$AS = h \cot 50^\circ = 72.445 \times 0.8391 = 60.78$ m

$BS = h \cot 35^\circ = 72.445 \times 1.4281 = 103.47$ m

$\tan(\angle SAB) = \frac{103.47}{60.78} = 1.7024$

$\angle SAB = \tan^{-1}(1.7024) = 59.57...^\circ$

Bearing of $B$ from $A = 270^\circ - (90^\circ - 59.57^\circ) = 270^\circ - 30.43^\circ$ —

More directly: From $A$, north is $000^\circ$. $S$ is at bearing $000^\circ$. $B$ is west of $S$, so from $A$, $B$ is at bearing $360^\circ - \angle SAB$ measured from north going clockwise... Actually, bearing of $B$ from $A$: $A$ is south of $S$, $B$ is west of $S$. So from $A$, looking north, $B$ is to the left (west) of north.

Bearing of $B$ from $A = 360^\circ - 59.6^\circ = 300.4^\circ$

Wait — let me reconsider. From point $A$, the direction to $S$ is due north ($000^\circ$). The angle between $AS$ and $AB$ is $\angle SAB = 59.6^\circ$. Since $B$ is to the west of $S$ (as $B$ is due west of $S$), from $A$, $B$ is to the left of north.

Bearing of $B$ from $A = 360^\circ - 59.6^\circ = 300.4^\circ$

$\boxed{\text{Bearing of } B \text{ from } A = 300^\circ}$ (nearest degree)

Marking: M1 for finding angle SAB; A1 for bearing 300°.

(e) [2 marks]

The angle of depression of $A$ from the top $T$ of the tower equals the angle of elevation of $T$ from $A$, which is given as $50^\circ$.

$\boxed{\text{Angle of depression} = 50^\circ}$

Marking: M1 for understanding angle of depression = angle of elevation; A1 for 50°.

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17. [5 marks total]

(a) [1 mark]

In $\triangle ABC$, $\angle C = 90^\circ$, $AC = 12$ cm, $BC = 5$ cm.

$AB^2 = AC^2 + BC^2 = 12^2 + 5^2 = 144 + 25 = 169$

$AB = \sqrt{169} = 13$

$\boxed{AB = 13 \text{ cm}}$

Mark for 13 cm.

(b) [2 marks]

Area of $\triangle ABC = \frac{1}{2} \times AC \times BC = \frac{1}{2} \times 12 \times 5 = 30$ cm²

Also, area $= \frac{1}{2} \times AB \times CD = \frac{1}{2} \times 13 \times CD$

$\frac{1}{2} \times 13 \times CD = 30$

$CD = \frac{60}{13} = 4.615...$

$\boxed{CD = 4.6 \text{ cm}}$ (to 1 d.p.)

Marking: M1 for area = 30 and setting up equation; A1 for 4.6 cm.

(c) [2 marks]

In right-angled triangle $ACD$ (since $CD \perp AB$):

$\tan(\angle ACD) = \frac{AD}{CD}$

First find $AD$: In $\triangle ACD$, $AC^2 = AD^2 + CD^2$

$12^2 = AD^2 + \left(\frac{60}{13}\right)^2$

$144 = AD^2 + \frac{3600}{169}$

$AD^2 = 144 - \frac{3600}{169} = \frac{24336 - 3600}{169} = \frac{20736}{169}$

$AD = \frac{144}{13}$ cm

$\tan(\angle ACD) = \frac{AD}{CD} = \frac{144/13}{60/13} = \frac{144}{60} = \frac{12}{5} = 2.4$

$\angle ACD = \tan^{-1}(2.4) = 67.380...^\circ$

$\boxed{\angle ACD = 67.4^\circ}$

Marking: M1 for finding AD and setting up tan ratio; A1 for 67.4°.

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18. [4 marks total]

(a) [2 marks]

Let $d_1$ = distance of car 1 (angle of depression $18^\circ$) from base. Let $d_2$ = distance of car 2 (angle of depression $32^\circ$) from base.

$\tan 18^\circ = \frac{60}{d_1} \Rightarrow d_1 = \frac{60}{\tan 18^\circ} = \frac{60}{0.3249} = 184.66...$

$\tan 32^\circ = \frac{60}{d_2} \Rightarrow d_2 = \frac{60}{\tan 32^\circ} = \frac{60}{0.6249} = 96.02...$

$\boxed{d_1 = 184.7 \text{ m}, \quad d_2 = 96.0 \text{ m}}$

Marking: 1 mark each for correct distances.

(b) [2 marks]

Distance between the two cars $= d_1 - d_2 = 184.66 - 96.02 = 88.64...$

$\boxed{88.6 \text{ m}}$

Marking: M1 for subtracting; A1 for 88.6 m.

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19. [4 marks total]

(a) [2 marks]

In $\triangle PQR$, $\angle P = 90^\circ$, $PQ = 45$ m, $\angle PRQ = 22^\circ$.

$\tan 22^\circ = \frac{PQ}{PR}$

$PR = \frac{PQ}{\tan 22^\circ} = \frac{45}{0.4040} = 111.38...$

$\boxed{PR = 111.4 \text{ m}}$

Marking: M1 for correct tan setup; A1 for 111.4 m.

(b) [2 marks]

First find $QR$: $\sin 22^\circ = \frac{PQ}{QR}$

$QR = \frac{45}{\sin 22^\circ} = \frac{45}{0.3746} = 120.13...$

Area $= \frac{1}{2} \times PQ \times PR = \frac{1}{2} \times 45 \times 111.38 = 2506.05...$

$\boxed{\text{Area} = 2506.1 \text{ m}^2}$

Marking: M1 for area formula with correct values; A1 for 2506.1 m².

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20. [6 marks total]

(a) [2 marks]

Bearing of $Y$ from $X$ is $048^\circ$. Bearing of $Z$ from $Y$ is $138^\circ$.

At $Y$, the back-bearing of $X$ from $Y$ is $048^\circ + 180^\circ = 228^\circ$.

$\angle XYZ = 228^\circ - 138^\circ = 90^\circ$

$\boxed{\angle XYZ = 90^\circ}$

Marking: M1 for back-bearing; A1 for 90°.

(b) [2 marks]

Since $\angle XYZ = 90^\circ$:

$XZ^2 = XY^2 + YZ^2 = 350^2 + 280^2 = 122500 + 78400 = 200900$

$XZ = \sqrt{200900} = 448.218...$

$\boxed{XZ = 448.2 \text{ km}}$

Marking: M1 for Pythagoras; A1 for 448.2 km.

(c) [2 marks]

We need the bearing of $X$ from $Z$.

In $\triangle XYZ$, $\tan(\angle XZY) = \frac{XY}{YZ} = \frac{350}{280} = 1.25$

$\angle XZY = \tan^{-1}(1.25) = 51.340...^\circ$

At $Z$, the bearing of $Y$ from $Z$ is $138^\circ + 180^\circ = 318^\circ$.

The bearing of $X$ from $Z$: From direction $ZY$ (bearing $318^\circ$), we rotate towards $X$. Since $X$ is "to the left" when facing from $Z$ to $Y$ (as the triangle goes $X \to Y \to Z$ with a right angle at $Y$), we subtract the angle.

Bearing of $X$ from $Z = 318^\circ - 51.34^\circ = 266.66...^\circ$

$\boxed{\text{Bearing of } X \text{ from } Z = 267^\circ}$ (nearest degree)

Marking: M1 for finding angle XZY and setting up bearing calculation; A1 for 267°.

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