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

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TuitionGoWhere Practice Paper - Elementary Mathematics Secondary 3

TuitionGoWhere Secondary School (AI)
Assessment: SA2 Practice Paper (Version 5)
Subject: Elementary Mathematics
Level: Secondary 3
Paper: SA2 Practice (Geometry & Trigonometry Focus)
Duration: 1 hour 30 minutes
Total Marks: 80

Name: __________________________
Class: __________________________
Date: __________________________

Instructions to Candidates

Write your name, class, and date in the spaces provided.
Answer all questions.
Write your answers in the spaces provided in this booklet.
If working is needed for any question, it must be shown below that question.
The number of marks is given in brackets [ ] at the end of each question or part question.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
Take $\pi$ to be $3.142$ or use the $\pi$ button on your calculator.

Section A: Basic Concepts and Calculations [30 Marks]

1. In the right-angled triangle $ABC$ , angle $ABC = 90^\circ$ . $AB = 7$ cm and $BC = 10$ cm.
Calculate the length of $AC$ .
[2]

Answer: __________________________ cm

2. In triangle $PQR$ , $PQ = 12$ cm, $PR = 15$ cm, and angle $QPR = 40^\circ$ .
Calculate the area of triangle $PQR$ .
[2]

Answer: __________________________ cm $^2$

3. Convert $2.4$ radians to degrees. Give your answer correct to 1 decimal place.
[2]

Answer: __________________________ $^\circ$

4. A sector of a circle has radius $8$ cm and an angle of $1.2$ radians.
Calculate the arc length of this sector.
[2]

Answer: __________________________ cm

5. In triangle $XYZ$ , $XY = 9$ cm, $YZ = 11$ cm, and $XZ = 14$ cm.
Calculate the size of angle $XYZ$ .
[3]

Answer: __________________________ $^\circ$

6. The bearing of point $A$ from point $B$ is $135^\circ$ .
Find the bearing of point $B$ from point $A$ .
[2]

Answer: __________________________ $^\circ$

7. Solve the equation $\sin \theta = 0.6$ for $0^\circ \le \theta \le 360^\circ$ .
[2]

Answer: $\theta =$ __________________________ $^\circ$ and __________________________ $^\circ$

8. A ladder of length $5$ m leans against a vertical wall. The foot of the ladder is $1.5$ m from the base of the wall.
Calculate the angle the ladder makes with the horizontal ground.
[2]

Answer: __________________________ $^\circ$

9. In a circle with centre $O$ and radius $10$ cm, a chord $AB$ has length $12$ cm.
Calculate the perpendicular distance from $O$ to the chord $AB$ .
[3]

Answer: __________________________ cm

10. Given that $\cos \alpha = -\frac{3}{5}$ and $90^\circ < \alpha < 180^\circ$ , find the exact value of $\sin \alpha$ .
[2]

Answer: $\sin \alpha =$ __________________________

Section B: Structured Problems and Applications [30 Marks]

11. The diagram shows a cuboid $ABCDEFGH$ .
$AB = 8$ cm, $BC = 6$ cm, and $CG = 5$ cm.
$M$ is the midpoint of $BC$ .

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

Answer: __________________________ cm

(b) Calculate the length of $GM$ .
[2]

Answer: __________________________ cm

Answer: __________________________ $^\circ$

12. Triangle $ABC$ has sides $AB = c$ , $BC = a$ , and $AC = b$ .
Given $a = 7$ cm, $b = 9$ cm, and angle $C = 60^\circ$ .

(a) Use the Cosine Rule to calculate the length of side $c$ ( $AB$ ).
[3]

Answer: __________________________ cm

(b) Hence, or otherwise, calculate the area of triangle $ABC$ .
[2]

Answer: __________________________ cm $^2$

13. Points $A$ , $B$ , and $C$ lie on a horizontal plane.
The bearing of $B$ from $A$ is $050^\circ$ .
The bearing of $C$ from $B$ is $140^\circ$ .
$AB = 100$ m and $BC = 80$ m.

(a) Calculate angle $ABC$ .
[2]

Answer: __________________________ $^\circ$

(b) Calculate the distance $AC$ .
[3]

Answer: __________________________ m

Answer: __________________________ $^\circ$

14. The diagram shows a circle with centre $O$ . Points $A$ , $B$ , and $C$ lie on the circumference.
$AC$ is a diameter. Angle $OAB = 35^\circ$ .

(a) State the reason why angle $ABC = 90^\circ$ .
[1]

Answer: _________________________________________________________________

(b) Calculate angle $ACB$ .
[2]

Answer: __________________________ $^\circ$

(c) If the radius of the circle is $6$ cm, calculate the length of arc $AB$ (the minor arc). Give your answer in terms of $\pi$ .
[2]

Answer: __________________________ cm

15. A sector $OAB$ has radius $r$ cm and angle $\theta$ radians.
The area of the sector is $20$ cm $^2$ and the arc length is $8$ cm.

(a) Write down two equations connecting $r$ and $\theta$ based on the formulas for area and arc length.
[2]

Answer:

(b) Solve these equations to find the values of $r$ and $\theta$ .
[3]

Answer: $r =$ __________________________ cm, $\theta =$ __________________________ rad

Section C: Complex Reasoning and Synthesis [20 Marks]

16. The diagram shows a triangular prism $ABCDEF$ .
The cross-section $ABC$ is an isosceles triangle with $AB = AC = 10$ cm and $BC = 12$ cm.
The length of the prism is $15$ cm.
$M$ is the midpoint of $BC$ .

(a) Calculate the height $AM$ of triangle $ABC$ .
[2]

Answer: __________________________ cm

(b) Calculate the angle between the plane $ABC$ and the plane $BCFE$ .
[1]

Answer: __________________________ $^\circ$

Answer: __________________________ $^\circ$

17. A vertical tower $TP$ stands on horizontal ground. Points $A$ and $B$ are on the ground such that $A$ , $B$ , and $P$ are collinear, with $B$ between $A$ and $P$ .
The angle of elevation of the top of the tower $T$ from $A$ is $25^\circ$ .
The angle of elevation of $T$ from $B$ is $40^\circ$ .
The distance $AB = 50$ m.

(a) Show that the height $h$ of the tower is given by $h = \frac{50}{\cot 25^\circ - \cot 40^\circ}$ .
[3]

Answer: (Show working)

(b) Calculate the height of the tower.
[2]

Answer: __________________________ m

Answer: __________________________ m

18. In triangle $PQR$ , $PQ = 8$ cm, $QR = 10$ cm, and angle $PQR = 120^\circ$ .

(a) Calculate the length of $PR$ .
[3]

Answer: __________________________ cm

(b) Calculate the area of triangle $PQR$ .
[2]

Answer: __________________________ cm $^2$

Answer: __________________________ cm

19. A circle has centre $O$ and radius $5$ cm. A tangent $PT$ touches the circle at $T$ . $OP = 13$ cm.

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

Answer: __________________________ cm

(b) Calculate angle $TOP$ .
[2]

Answer: __________________________ $^\circ$

(c) Calculate the area of the shaded region bounded by the tangent $PT$ , the line $OP$ , and the arc $TQ$ (where $Q$ is the intersection of $OP$ and the circle).
[3]

Answer: __________________________ cm $^2$

20. The function $f(x) = 3 \sin(2x) + 1$ is defined for $0^\circ \le x \le 360^\circ$ .

(a) State the amplitude and period of the function.
[2]

Answer: Amplitude = __________________________, Period = __________________________ $^\circ$

(b) Solve the equation $3 \sin(2x) + 1 = 2.5$ for $0^\circ \le x \le 360^\circ$ .
[4]

Answer: $x =$ __________________________ $^\circ$ , __________________________ $^\circ$ , __________________________ $^\circ$ , __________________________ $^\circ$

End of Paper

Answers

TuitionGoWhere Practice Paper - Elementary Mathematics Secondary 3

Answer Key and Marking Scheme (Version 5)

Subject: Elementary Mathematics
Level: Secondary 3
Topic: Geometry & Trigonometry

Section A: Basic Concepts and Calculations

1.
Using Pythagoras' Theorem:
$AC^2 = AB^2 + BC^2$
$AC^2 = 7^2 + 10^2 = 49 + 100 = 149$
$AC = \sqrt{149} \approx 12.206$
Answer: $12.2$ cm [2]
(1 mark for substitution, 1 mark for correct answer)

2.
Area $= \frac{1}{2} ab \sin C$
Area $= \frac{1}{2} (12)(15) \sin 40^\circ$
Area $= 90 \times 0.64278...$
Area $\approx 57.85$
Answer: $57.9$ cm $^2$ [2]
(1 mark for formula/substitution, 1 mark for answer)

3.
Degrees $= \text{Radians} \times \frac{180}{\pi}$
Degrees $= 2.4 \times \frac{180}{\pi} \approx 137.509...$
Answer: $137.5^\circ$ [2]
(1 mark for conversion factor, 1 mark for answer)

4.
Arc length $s = r\theta$
$s = 8 \times 1.2 = 9.6$
Answer: $9.6$ cm [2]

5.
Using Cosine Rule: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$
Here $a=11, c=9, b=14$ .
$\cos(\angle XYZ) = \frac{11^2 + 9^2 - 14^2}{2(11)(9)}$
$\cos(\angle XYZ) = \frac{121 + 81 - 196}{198} = \frac{6}{198} = \frac{1}{33}$
$\angle XYZ = \cos^{-1}(\frac{1}{33}) \approx 88.26^\circ$
Answer: $88.3^\circ$ [3]
(1 mark for formula, 1 mark for substitution, 1 mark for answer)

6.
Back bearing $= 135^\circ + 180^\circ = 315^\circ$
Answer: $315^\circ$ [2]

7.
Reference angle $\alpha = \sin^{-1}(0.6) \approx 36.87^\circ$
Sine is positive in 1st and 2nd quadrants.
$\theta_1 = 36.87^\circ$
$\theta_2 = 180^\circ - 36.87^\circ = 143.13^\circ$
Answer: $36.9^\circ$ and $143.1^\circ$ [2]
(1 mark for each correct angle)

8.
Let angle be $\theta$ .
$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{1.5}{5} = 0.3$
$\theta = \cos^{-1}(0.3) \approx 72.54^\circ$
Answer: $72.5^\circ$ [2]

9.
Let $M$ be the midpoint of $AB$ . $AM = 6$ cm.
Triangle $OMA$ is right-angled at $M$ .
$OM^2 + AM^2 = OA^2$
$OM^2 + 6^2 = 10^2$
$OM^2 = 100 - 36 = 64$
$OM = 8$
Answer: $8$ cm [3]
(1 mark for identifying right triangle/half-chord, 1 mark for Pythagoras setup, 1 mark for answer)

10.
$\sin^2 \alpha + \cos^2 \alpha = 1$
$\sin^2 \alpha + (-\frac{3}{5})^2 = 1$
$\sin^2 \alpha + \frac{9}{25} = 1$
$\sin^2 \alpha = \frac{16}{25}$
$\sin \alpha = \pm \frac{4}{5}$
Since $90^\circ < \alpha < 180^\circ$ (2nd quadrant), sine is positive.
Answer: $\frac{4}{5}$ [2]
(1 mark for magnitude, 1 mark for correct sign)

Section B: Structured Problems and Applications

11.
(a) $M$ is midpoint of $BC$ ( $6$ cm).
$BM = \frac{6}{2} = 3$ cm.
Answer: $3$ cm [1]

(b) In $\triangle GCM$ (right-angled at $C$ ):
$GC = 5$ cm, $CM = 3$ cm.
$GM^2 = 5^2 + 3^2 = 25 + 9 = 34$
$GM = \sqrt{34} \approx 5.83$
Answer: $5.83$ cm [2]

(c) The angle between line $GM$ and base $ABCD$ is angle $GMC$ .
In $\triangle GCM$ , $\tan(\angle GMC) = \frac{GC}{CM} = \frac{5}{3}$
$\angle GMC = \tan^{-1}(\frac{5}{3}) \approx 59.036^\circ$
Answer: $59.0^\circ$ [3]
(1 mark for identifying angle, 1 mark for trig ratio, 1 mark for answer)

12.
(a) $c^2 = a^2 + b^2 - 2ab \cos C$
$c^2 = 7^2 + 9^2 - 2(7)(9) \cos 60^\circ$
$c^2 = 49 + 81 - 126(0.5)$
$c^2 = 130 - 63 = 67$
$c = \sqrt{67} \approx 8.185$
Answer: $8.19$ cm [3]

(b) Area $= \frac{1}{2} ab \sin C$
Area $= \frac{1}{2} (7)(9) \sin 60^\circ$
Area $= 31.5 \times \frac{\sqrt{3}}{2} \approx 27.28$
Answer: $27.3$ cm $^2$ [2]

13.
(a) Bearing of $B$ from $A$ is $050^\circ$ . North lines are parallel.
Angle at $B$ (inside triangle) relative to North:
Back bearing of $A$ from $B$ is $050^\circ + 180^\circ = 230^\circ$ .
Angle $ABC = 230^\circ - 140^\circ = 90^\circ$ .
Alternatively: Co-interior angles sum to $180^\circ$ . Angle between $AB$ and South at $B$ is $50^\circ$ . Angle between $BC$ and North at $B$ is $180-140=40$ ? No.
Let's use geometry:
North at $B$ . Line $BA$ is $180+50 = 230^\circ$ bearing. Line $BC$ is $140^\circ$ bearing.
Angle $ABC = 230^\circ - 140^\circ = 90^\circ$ .
Answer: $90^\circ$ [2]

(b) Since $\triangle ABC$ is right-angled at $B$ :
$AC^2 = AB^2 + BC^2 = 100^2 + 80^2 = 10000 + 6400 = 16400$
$AC = \sqrt{16400} \approx 128.06$
Answer: $128$ m [3]

(c) In right $\triangle ABC$ :
$\tan(\angle BCA) = \frac{AB}{BC} = \frac{100}{80} = 1.25$
$\angle BCA = \tan^{-1}(1.25) \approx 51.34^\circ$
Bearing of $C$ from $B$ is $140^\circ$ .
North line at $C$ . Back bearing of $B$ from $C$ is $140^\circ + 180^\circ = 320^\circ$ .
Bearing of $A$ from $C$ = Back bearing of $B$ from $C$ + $\angle BCA$ ?
Let's visualize. $B$ is North-East of $A$ ? No, $B$ is $050$ from $A$ . $C$ is $140$ from $B$ .
Triangle is right angled at $B$ .
Bearing $C$ to $B$ is $320^\circ$ .
Angle $BCA$ is $51.3^\circ$ . $A$ is to the "left" of line $CB$ when standing at $C$ looking at $B$ ?
Vector $CB$ is bearing $320$ . Vector $CA$ is rotated counter-clockwise by $51.3$ ?
Let's check coordinates.
$A=(0,0)$ . $B=(100\sin50, 100\cos50) \approx (76.6, 64.3)$ .
$C = B + (80\sin140, 80\cos140) \approx (76.6+51.4, 64.3-51.4) = (128, 12.9)$ .
Vector $CA = A - C = (-128, -12.9)$ .
Angle $\theta = \tan^{-1}(\frac{-128}{-12.9})$ . Both negative -> 3rd quadrant.
Ref angle $\tan^{-1}(128/12.9) \approx 84.2^\circ$ .
Bearing $= 180 + 84.2 = 264.2^\circ$ .
Let's re-evaluate geometric addition.
Bearing $B$ from $C$ is $320^\circ$ .
Angle $BCA = 51.3^\circ$ .
Is $A$ clockwise or anti-clockwise from $B$ relative to $C$ ?
$A$ is West of $C$ . $B$ is North-West of $C$ .
So $A$ is clockwise from $B$ ? No.
Bearing $C \to B$ is $320$ .
Angle $BCA$ is inside the triangle.
Bearing $C \to A = 320^\circ + 51.3^\circ = 371.3^\circ \equiv 11.3^\circ$ ? No.
Let's stick to coordinates for safety in marking.
$\Delta x = -128$ , $\Delta y = -12.9$ .
$\tan \alpha = 128/12.9$ . $\alpha = 84.2^\circ$ .
Since $\Delta x < 0, \Delta y < 0$ , it is in 3rd quadrant relative to C?
Wait, $A$ is origin. $C$ is $(128, 12.9)$ .
Vector $CA$ is $(-128, -12.9)$ .
Angle from North (positive y):
Standard angle from positive x-axis: $180 + \tan^{-1}(12.9/128) \approx 185.7^\circ$ .
Bearing is clockwise from North (positive y).
North is $90^\circ$ in standard math angle? No, North is $0^\circ$ bearing.
Let's use bearing logic.
North at $C$ . Line $CB$ is bearing $320^\circ$ .
Line $CA$ ?
Angle of $CB$ with North is $40^\circ$ to the Left (West).
Angle $BCA = 51.3^\circ$ .
So $CA$ is $51.3 - 40 = 11.3^\circ$ to the Right (East) of South?
Let's use the coordinate result:
$\tan^{-1}(128/12.9) = 84.2^\circ$ from Vertical (South).
Since $x$ is negative (West) and $y$ is negative (South), it is South-West.
Bearing $= 180^\circ + 84.2^\circ = 264.2^\circ$ .
Answer: $264^\circ$ [3]
(1 mark for angle BCA, 1 mark for bearing logic, 1 mark for answer)

14.
(a) Angle in a semicircle is $90^\circ$ . [1]
(b) In $\triangle ABC$ , angle $B=90^\circ$ , angle $A=35^\circ$ .
Angle $C = 180 - 90 - 35 = 55^\circ$ .
Answer: $55^\circ$ [2]
(c) Angle at centre $AOB = 2 \times$ Angle at circumference $ACB$ ? No.
Triangle $OAB$ is isosceles ( $OA=OB$ ). Angle $OAB = 35^\circ$ , so Angle $OBA = 35^\circ$ .
Angle $AOB = 180 - 35 - 35 = 110^\circ$ .
Convert to radians: $110 \times \frac{\pi}{180} = \frac{11\pi}{18}$ .
Arc length $= r\theta = 6 \times \frac{11\pi}{18} = \frac{11\pi}{3}$ .
Answer: $\frac{11\pi}{3}$ cm [2]

15.
(a) Area $= \frac{1}{2}r^2\theta = 20 \Rightarrow r^2\theta = 40$ .
Arc length $= r\theta = 8$ .
Answer: Equations stated. [2]
(b) From (2), $\theta = \frac{8}{r}$ .
Substitute into (1): $r^2(\frac{8}{r}) = 40 \Rightarrow 8r = 40 \Rightarrow r = 5$ .
$\theta = \frac{8}{5} = 1.6$ .
Answer: $r=5$ cm, $\theta=1.6$ rad [3]

Section C: Complex Reasoning and Synthesis

16.
(a) $M$ is midpoint of $BC$ , so $BM = 6$ cm.
In $\triangle ABM$ (right-angled at $M$ ):
$AM^2 + 6^2 = 10^2 \Rightarrow AM^2 = 100 - 36 = 64 \Rightarrow AM = 8$ cm.
Answer: $8$ cm [2]

(b) The prism is a right prism, so the side faces are perpendicular to the base.
However, the question asks for angle between plane $ABC$ and plane $BCFE$ .
Plane $BCFE$ is a vertical rectangular face. Plane $ABC$ is the triangular base?
No, usually "base" refers to the face it rests on. If it rests on $BCFE$ , then $ABC$ is a vertical cross section?
Standard orientation: $ABC$ is cross section. $BCFE$ is a rectangular face.
The angle between the triangular face $ABC$ and the rectangular base $BCFE$ ?
If the prism lies on face $BCFE$ , then the angle is the angle between $AM$ and the plane $BCFE$ .
Since $AM \perp BC$ and the face $BCFE$ is perpendicular to the plane containing $AM$ ?
Actually, in a standard right prism, the lateral faces are perpendicular to the cross-section.
So the angle between plane $ABC$ and plane $BCFE$ is $90^\circ$ .
Answer: $90^\circ$ [1]

(c) Angle between line $AF$ and base plane $BCFE$ .
Projection of $A$ onto plane $BCFE$ is $M$ (since $AM \perp BC$ and $AM \perp$ vertical edges? No. $AM$ is in the plane of the triangle. The triangle is perpendicular to the length.
So $AM$ is perpendicular to the face $BCFE$ ? Yes, if $ABC$ is the cross section and $BCFE$ is a lateral face?
Wait. $BCFE$ contains edge $BC$ . $AM$ is altitude to $BC$ .
Since the prism is right, the plane $ABC$ is perpendicular to the edges $AD, BE, CF$ .
Is $AM$ perpendicular to the plane $BCFE$ ?
$AM \perp BC$ . Is $AM \perp BE$ ? Yes, because $BE$ is perpendicular to the whole plane $ABC$ .
So $AM$ is perpendicular to the plane $BCFE$ .
Therefore, $M$ is the projection of $A$ onto the plane $BCFE$ .
The angle is $\angle AFM$ .
In $\triangle AMF$ (right-angled at $M$ ):
$AM = 8$ cm.
$MF$ is the diagonal of the base rectangle? No. $F$ is a vertex. $M$ is on $BC$ .
$BCFE$ is a rectangle $12 \times 15$ . $M$ is midpoint of $BC$ .
$F$ is corner opposite $B$ ? $B-C-F-E$ ? No, $B-C$ is width. $C-F$ is length.
So $M$ is on $BC$ . $F$ is at corner.
Distance $MF$ : In rectangle $BCFE$ , $M$ is mid $BC$ . $F$ is vertex.
$\triangle MCF$ is right angled at $C$ .
$MC = 6$ cm. $CF = 15$ cm (length of prism).
$MF^2 = 6^2 + 15^2 = 36 + 225 = 261$ .
$MF = \sqrt{261} \approx 16.155$ cm.
In $\triangle AMF$ : $\tan(\angle AFM) = \frac{AM}{MF} = \frac{8}{\sqrt{261}}$ .
$\angle AFM = \tan^{-1}(\frac{8}{16.155}) \approx 26.35^\circ$ .
Answer: $26.4^\circ$ [4]
(1 mark for identifying projection M, 1 mark for length MF, 1 mark for trig ratio, 1 mark for answer)

17.
(a) Let $BP = x$ . Then $AP = x + 50$ .
In $\triangle TBP$ : $\tan 40^\circ = \frac{h}{x} \Rightarrow x = h \cot 40^\circ$ .
In $\triangle TAP$ : $\tan 25^\circ = \frac{h}{x+50} \Rightarrow x+50 = h \cot 25^\circ$ .
Subtracting: $(x+50) - x = h \cot 25^\circ - h \cot 40^\circ$ .
$50 = h(\cot 25^\circ - \cot 40^\circ)$ .
$h = \frac{50}{\cot 25^\circ - \cot 40^\circ}$ . [3]

(b) $h = \frac{50}{2.1445 - 1.1917} = \frac{50}{0.9528} \approx 52.47$
Answer: $52.5$ m [2]

18.
(a) $PR^2 = 8^2 + 10^2 - 2(8)(10)\cos 120^\circ$ .
$\cos 120^\circ = -0.5$ .
$PR^2 = 64 + 100 - 160(-0.5) = 164 + 80 = 244$ .
$PR = \sqrt{244} \approx 15.62$
Answer: $15.6$ cm [3]

(b) Area $= \frac{1}{2}(8)(10)\sin 120^\circ = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \approx 34.64$
Answer: $34.6$ cm $^2$ [2]

(c) Area $= \frac{1}{2} \times \text{base} \times \text{height}$ .
$34.64 = \frac{1}{2} (15.62) (QS)$ .
$QS = \frac{2 \times 34.64}{15.62} \approx 4.436$
Answer: $4.44$ cm [2]

19.
(a) $\triangle OTP$ is right-angled at $T$ (tangent $\perp$ radius).
$PT^2 + OT^2 = OP^2$ .
$PT^2 + 5^2 = 13^2 \Rightarrow PT^2 = 169 - 25 = 144$ .
$PT = 12$ cm.
Answer: $12$ cm [2]

(b) $\cos(\angle TOP) = \frac{OT}{OP} = \frac{5}{13}$ .
$\angle TOP = \cos^{-1}(\frac{5}{13}) \approx 67.38^\circ$ .
Answer: $67.4^\circ$ [2]

(c) Area of $\triangle OTP = \frac{1}{2}(5)(12) = 30$ cm $^2$ .
Area of Sector $OTQ$ (angle $67.38^\circ$ ):
Angle in rad $= 67.38 \times \frac{\pi}{180} \approx 1.176$ rad.
Area Sector $= \frac{1}{2} r^2 \theta = \frac{1}{2}(25)(1.176) \approx 14.70$ cm $^2$ .
Shaded Area $= 30 - 14.70 = 15.30$ cm $^2$ .
Answer: $15.3$ cm $^2$ [3]

20.
(a) Amplitude $= 3$ .
Period $= \frac{360^\circ}{2} = 180^\circ$ .
Answer: Amp $3$ , Period $180^\circ$ [2]

(b) $3 \sin(2x) + 1 = 2.5 \Rightarrow 3 \sin(2x) = 1.5 \Rightarrow \sin(2x) = 0.5$ .
Let $u = 2x$ . Range for $u$ : $0^\circ \le u \le 720^\circ$ .
Basic angle for $\sin u = 0.5$ is $30^\circ$ .
Solutions for $u$ :
$u_1 = 30^\circ$
$u_2 = 180 - 30 = 150^\circ$
$u_3 = 360 + 30 = 390^\circ$
$u_4 = 540 - 30 = 510^\circ$
$2x = 30, 150, 390, 510$ .
$x = 15, 75, 195, 255$ .
Answer: $15^\circ, 75^\circ, 195^\circ, 255^\circ$ [4]
(1 mark for basic angle, 1 mark for all 4 u values, 1 mark for dividing by 2, 1 mark for final list)

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