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Secondary 3 Additional Mathematics Geometry Trigonometry Quiz

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Secondary 3 Additional Mathematics Quiz - Geometry Trigonometry

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: ________ / 60

Duration: 45 minutes
Total Marks: 60

Instructions:

Answer ALL questions.
Show all working clearly. Marks are awarded for correct reasoning and method, not only for the final answer.
Non-exact answers should be given correct to 3 significant figures unless otherwise stated.
The use of a scientific calculator is allowed.
Diagrams are not drawn to scale unless otherwise indicated.

Section A: Trigonometric Identities and Equations (Questions 1–5)

Questions 1–5 carry 2 marks each.

1. Express $\frac{\sin^2 \theta}{1 - \cos \theta}$ in terms of $\cos \theta$ only, and hence simplify the expression completely.

2. Solve the equation $2\cos^2 x - 3\cos x + 1 = 0$ for $0^\circ \leq x \leq 360^\circ$ .

3. Prove the identity: $\sec^2 \theta - \tan^2 \theta = 1$ .

4. Given that $\sin A = \frac{3}{5}$ and angle $A$ is acute, find the exact value of $\cos 2A$ .

5. Solve the equation $\tan 2x = 1$ for $0^\circ \leq x \leq 180^\circ$ .

Section B: Coordinate Geometry — Straight Lines and Circles (Questions 6–12)

Questions 6–8 carry 3 marks each. Questions 9–12 carry 4 marks each.

6. The points $A(1, 3)$ and $B(5, 7)$ are given. Find the equation of the perpendicular bisector of the line segment $AB$ .

7. Find the coordinates of the centre and the radius of the circle with equation
$x^2 + y^2 - 6x + 4y - 12 = 0.$

8. The line $y = 2x + k$ is tangent to the circle $x^2 + y^2 = 25$ . Find the possible values of $k$ .

9. A circle has centre $C(2, -1)$ and passes through the point $P(5, 3)$ .

(a) Find the equation of the circle. (2 marks)

(b) Show that the point $Q(-1, -5)$ lies on the circle. (1 mark)

10. The line $l_1$ has equation $3x - 4y + 8 = 0$ . The line $l_2$ passes through the point $(6, -2)$ and is perpendicular to $l_1$ .

(a) Find the equation of $l_2$ . (2 marks)

(b) Find the coordinates of the point of intersection of $l_1$ and $l_2$ . (2 marks)

11. The points $A(-2, 1)$ , $B(4, 5)$ , and $C(6, -1)$ form a triangle.

(a) Find the length of $AB$ . (1 mark)

(b) Find the equation of the median from $C$ to the midpoint of $AB$ . (3 marks)

12. A circle has equation $(x - 3)^2 + (y + 2)^2 = 20$ . The line $y = x - 1$ intersects the circle at two points $P$ and $Q$ .

(a) Find the coordinates of $P$ and $Q$ . (3 marks)

(b) Find the exact length of the chord $PQ$ . (1 mark)

Section C: Trigonometry — Graphs, R-Formula, and Applications (Questions 13–20)

Questions 13–15 carry 3 marks each. Questions 16–20 carry 4 marks each.

13. Given that $5\sin \theta + 12\cos \theta = R\sin(\theta + \alpha)$ , where $R > 0$ and $0^\circ < \alpha < 90^\circ$ , find the values of $R$ and $\alpha$ .

14. Sketch the graph of $y = 3\cos 2x$ for $0^\circ \leq x \leq 360^\circ$ , clearly indicating the amplitude, period, and all intercepts with the axes.

15. Solve the equation $\sin\left(x + 30^\circ\right) = \frac{1}{2}$ for $0^\circ \leq x \leq 360^\circ$ .

16. Express $4\cos x - 3\sin x$ in the form $R\cos(x + \alpha)$ , where $R > 0$ and $0^\circ < \alpha < 90^\circ$ . Hence find the maximum value of $4\cos x - 3\sin x$ and the value of $x$ at which it occurs for $0^\circ \leq x \leq 360^\circ$ .

17. In triangle $PQR$ , $PQ = 8$ cm, $QR = 11$ cm, and $\angle PQR = 52^\circ$ .

(a) Calculate the length of $PR$ , giving your answer correct to 3 significant figures. (2 marks)

(b) Calculate the area of triangle $PQR$ , giving your answer correct to 3 significant figures. (2 marks)

18. The diagram shows a triangle $ABC$ where $AB = 12$ cm, $AC = 9$ cm, and $\angle BAC = 65^\circ$ .

(a) Find the length of $BC$ . (2 marks)

(b) Given that $D$ lies on $AB$ such that $CD$ is perpendicular to $AB$ , find the length of $CD$ . (2 marks)

19. From a point $A$ on the ground, the angle of elevation to the top of a building is $35^\circ$ . From a point $B$ , which is 40 m further away from the building on the same horizontal line as $A$ , the angle of elevation is $22^\circ$ .

(a) By forming two equations, show that the height $h$ of the building satisfies
$h = \frac{40}{\cot 22^\circ - \cot 35^\circ}.$ (2 marks)

(b) Hence calculate the height of the building, giving your answer correct to 3 significant figures. (2 marks)

20. The figure shows a quadrilateral $ABCD$ where $AB = 6$ cm, $BC = 8$ cm, $CD = 5$ cm, $\angle ABC = 110^\circ$ , and $\angle BCD = 70^\circ$ .

(a) Find the length of diagonal $AC$ . (2 marks)

(b) Find the area of quadrilateral $ABCD$ . (2 marks)

END OF QUIZ

Answers

Secondary 3 Additional Mathematics Quiz - Geometry Trigonometry

Answer Key

Section A: Trigonometric Identities and Equations

1. Express $\frac{\sin^2 \theta}{1 - \cos \theta}$ in terms of $\cos \theta$ only, and hence simplify.

Working: $\frac{\sin^2 \theta}{1 - \cos \theta} = \frac{1 - \cos^2 \theta}{1 - \cos \theta} = \frac{(1 - \cos \theta)(1 + \cos \theta)}{1 - \cos \theta} = 1 + \cos \theta$

Answer: $1 + \cos \theta$

Marks: 2

M1: Use identity $\sin^2 \theta = 1 - \cos^2 \theta$ and factorise
A1: Correct simplified answer $1 + \cos \theta$

Common mistake: Students may try to divide term-by-term instead of factorising the difference of squares.

2. Solve $2\cos^2 x - 3\cos x + 1 = 0$ for $0^\circ \leq x \leq 360^\circ$ .

Working: Let $u = \cos x$ : $2u^2 - 3u + 1 = 0$
$(2u - 1)(u - 1) = 0$
$u = \frac{1}{2}$ or $u = 1$

When $\cos x = \frac{1}{2}$ : $x = 60^\circ$ or $x = 300^\circ$
When $\cos x = 1$ : $x = 0^\circ$ or $x = 360^\circ$

Answer: $x = 0^\circ, 60^\circ, 300^\circ, 360^\circ$

Marks: 2

M1: Factorise or use quadratic formula correctly to find $\cos x = \frac{1}{2}$ or $1$
A1: All four correct values in the given range

Common mistake: Forgetting $x = 0^\circ$ and $360^\circ$ when $\cos x = 1$ ; only giving one solution per case.

3. Prove: $\sec^2 \theta - \tan^2 \theta = 1$ .

Working: $\sec^2 \theta - \tan^2 \theta = \frac{1}{\cos^2 \theta} - \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{1 - \sin^2 \theta}{\cos^2 \theta} = \frac{\cos^2 \theta}{\cos^2 \theta} = 1$

Answer: Proved.

Marks: 2

M1: Express in terms of $\sin \theta$ and $\cos \theta$ and combine into single fraction
A1: Correctly simplify to 1

Common mistake: Starting with the identity to be proved and manipulating both sides simultaneously (circular reasoning). Work from one side only.

4. Given $\sin A = \frac{3}{5}$ and $A$ is acute, find $\cos 2A$ .

Working: Since $A$ is acute, $\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$

$\cos 2A = 1 - 2\sin^2 A = 1 - 2\left(\frac{9}{25}\right) = 1 - \frac{18}{25} = \frac{7}{25}$

Answer: $\frac{7}{25}$

Marks: 2

M1: Use correct double-angle formula $\cos 2A = 1 - 2\sin^2 A$ (or equivalent)
A1: Correct exact answer $\frac{7}{25}$

Common mistake: Using $\cos 2A = 2\cos^2 A - 1$ but incorrectly finding $\cos A$ (e.g., forgetting that $A$ is acute so $\cos A > 0$ ).

5. Solve $\tan 2x = 1$ for $0^\circ \leq x \leq 180^\circ$ .

Working: $\tan 2x = 1 \Rightarrow 2x = 45^\circ, 225^\circ, 405^\circ, 585^\circ$
(adding $180^\circ$ each time; $2x$ ranges from $0^\circ$ to $360^\circ$ )

$x = 22.5^\circ, 112.5^\circ, 202.5^\circ, 292.5^\circ$

But $0^\circ \leq x \leq 180^\circ$ , so $x = 22.5^\circ, 112.5^\circ$

Answer: $x = 22.5^\circ, 112.5^\circ$

Marks: 2

M1: Correctly find $2x = 45^\circ, 225^\circ$ (within $0^\circ \leq 2x \leq 360^\circ$ )
A1: Both correct values of $x$

Common mistake: Forgetting that $2x$ ranges up to $360^\circ$ (not $180^\circ$ ), so missing solutions. Also, including $x = 202.5^\circ$ which is outside the range.

Section B: Coordinate Geometry — Straight Lines and Circles

6. Find the perpendicular bisector of $AB$ where $A(1, 3)$ and $B(5, 7)$ .

Working: Midpoint of $AB$ : $\left(\frac{1+5}{2}, \frac{3+7}{2}\right) = (3, 5)$

Gradient of $AB$ : $m_{AB} = \frac{7-3}{5-1} = \frac{4}{4} = 1$

Gradient of perpendicular bisector: $m = -1$

Equation: $y - 5 = -1(x - 3)$
$y - 5 = -x + 3$
$x + y = 8$

Answer: $x + y = 8$ (or $y = -x + 8$ )

Marks: 3

M1: Correct midpoint
M1: Correct perpendicular gradient
A1: Correct equation in required form

7. Find centre and radius of $x^2 + y^2 - 6x + 4y - 12 = 0$ .

Working: Complete the square:
$x^2 - 6x + y^2 + 4y = 12$
$(x - 3)^2 - 9 + (y + 2)^2 - 4 = 12$
$(x - 3)^2 + (y + 2)^2 = 25$

Answer: Centre $(3, -2)$ , radius $5$

Marks: 3

M1: Correctly complete the square in $x$ and $y$
A1: Correct centre
A1: Correct radius

8. Find $k$ such that $y = 2x + k$ is tangent to $x^2 + y^2 = 25$ .

Working: Substitute: $x^2 + (2x + k)^2 = 25$
$x^2 + 4x^2 + 4kx + k^2 = 25$
$5x^2 + 4kx + (k^2 - 25) = 0$

For tangency, $\Delta = 0$ :
$(4k)^2 - 4(5)(k^2 - 25) = 0$
$16k^2 - 20k^2 + 500 = 0$
$-4k^2 + 500 = 0$
$k^2 = 125$
$k = \pm 5\sqrt{5}$

Answer: $k = 5\sqrt{5}$ or $k = -5\sqrt{5}$

Marks: 3

M1: Substitute and form quadratic in $x$
M1: Set discriminant $= 0$
A1: Both correct values of $k$

9. Circle with centre $C(2, -1)$ through $P(5, 3)$ .

(a) Equation of the circle:

Working: $r^2 = (5-2)^2 + (3-(-1))^2 = 9 + 16 = 25$

Answer: $(x - 2)^2 + (y + 1)^2 = 25$

(b) Show $Q(-1, -5)$ lies on the circle:

Working: LHS $= (-1 - 2)^2 + (-5 + 1)^2 = 9 + 16 = 25 =$ RHS ✓

Answer: Verified.

(c) Tangent at $P$ :

Working: Gradient of $CP$ : $\frac{3-(-1)}{5-2} = \frac{4}{3}$
Gradient of tangent: $-\frac{3}{4}$

Equation: $y - 3 = -\frac{3}{4}(x - 5)$
$4y - 12 = -3x + 15$
$3x + 4y = 27$

Answer: $3x + 4y = 27$

Marks: 4 (2+1+1)

(a) M1: Correct $r^2$ ; A1: Correct equation
(b) M1: Substitute and verify
(c) M1: Correct gradient of tangent; A1: Correct equation

10. Line $l_1: 3x - 4y + 8 = 0$ ; $l_2$ through $(6, -2)$ , perpendicular to $l_1$ .

(a) Equation of $l_2$ :

Working: Gradient of $l_1$ : $\frac{3}{4}$
Gradient of $l_2$ : $-\frac{4}{3}$

Equation: $y + 2 = -\frac{4}{3}(x - 6)$
$3y + 6 = -4x + 24$
$4x + 3y = 18$

Answer: $4x + 3y = 18$

(b) Intersection of $l_1$ and $l_2$ :

Working: $l_1: 3x - 4y = -8$ … (i)
$l_2: 4x + 3y = 18$ … (ii)

(i) $\times$ 3: $9x - 12y = -24$
(ii) $\times$ 4: $16x + 12y = 72$

Add: $25x = 48 \Rightarrow x = \frac{48}{25}$

From (ii): $3y = 18 - 4\left(\frac{48}{25}\right) = \frac{450 - 192}{25} = \frac{258}{25}$
$y = \frac{86}{25}$

Answer: $\left(\frac{48}{25}, \frac{86}{25}\right)$

Marks: 4 (2+2)

(a) M1: Correct perpendicular gradient; A1: Correct equation
(b) M1: Correct elimination/substitution; A1: Both coordinates correct

11. Triangle with $A(-2, 1)$ , $B(4, 5)$ , $C(6, -1)$ .

(a) Length of $AB$ :

Working: $AB = \sqrt{(4-(-2))^2 + (5-1)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$

Answer: $2\sqrt{13}$ units (or $\sqrt{52}$ units)

(b) Median from $C$ to midpoint of $AB$ :

Working: Midpoint of $AB$ : $\left(\frac{-2+4}{2}, \frac{1+5}{2}\right) = (1, 3)$

Gradient of median: $\frac{-1 - 3}{6 - 1} = \frac{-4}{5}$

Equation: $y - 3 = -\frac{4}{5}(x - 1)$
$5y - 15 = -4x + 4$
$4x + 5y = 19$

Answer: $4x + 5y = 19$

Marks: 4 (1+3)

(a) A1: Correct length
(b) M1: Correct midpoint; M1: Correct gradient; A1: Correct equation

12. Circle $(x - 3)^2 + (y + 2)^2 = 20$ ; line $y = x - 1$ .

(a) Coordinates of $P$ and $Q$ :

Working: Substitute: $(x - 3)^2 + (x - 1 + 2)^2 = 20$
$(x - 3)^2 + (x + 1)^2 = 20$
$x^2 - 6x + 9 + x^2 + 2x + 1 = 20$
$2x^2 - 4x + 10 = 20$
$2x^2 - 4x - 10 = 0$
$x^2 - 2x - 5 = 0$
$x = \frac{2 \pm \sqrt{4 + 20}}{2} = \frac{2 \pm \sqrt{24}}{2} = 1 \pm \sqrt{6}$

When $x = 1 + \sqrt{6}$ : $y = \sqrt{6}$
When $x = 1 - \sqrt{6}$ : $y = -\sqrt{6}$

Answer: $P(1 + \sqrt{6}, \sqrt{6})$ and $Q(1 - \sqrt{6}, -\sqrt{6})$

(b) Length of chord $PQ$ :

Working: $PQ = \sqrt{[(1+\sqrt{6})-(1-\sqrt{6})]^2 + [\sqrt{6}-(-\sqrt{6})]^2}$
$= \sqrt{(2\sqrt{6})^2 + (2\sqrt{6})^2} = \sqrt{24 + 24} = \sqrt{48} = 4\sqrt{3}$

Answer: $4\sqrt{3}$ units

Marks: 4 (3+1)

(a) M1: Substitute and form quadratic; M1: Solve for $x$ ; A1: Both coordinates
(b) A1: Correct length

Section C: Trigonometry — Graphs, R-Formula, and Applications

13. Find $R$ and $\alpha$ for $5\sin \theta + 12\cos \theta = R\sin(\theta + \alpha)$ .

Working: $R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$

Comparing: $R\cos\alpha = 5$ , $R\sin\alpha = 12$

$R = \sqrt{25 + 144} = \sqrt{169} = 13$

$\tan\alpha = \frac{12}{5} \Rightarrow \alpha = \tan^{-1}\left(\frac{12}{5}\right) \approx 67.4^\circ$

Answer: $R = 13$ , $\alpha \approx 67.4^\circ$

Marks: 3

M1: Correct expansion of $R\sin(\theta + \alpha)$ and comparison
A1: $R = 13$
A1: $\alpha \approx 67.4^\circ$

14. Sketch $y = 3\cos 2x$ for $0^\circ \leq x \leq 360^\circ$ .

Answer:

Amplitude: $3$
Period: $\frac{360^\circ}{2} = 180^\circ$
$x$ -intercepts: $45^\circ, 135^\circ, 225^\circ, 315^\circ$
Maximum value $3$ at $x = 0^\circ, 180^\circ, 360^\circ$
Minimum value $-3$ at $x = 90^\circ, 270^\circ$

Marks: 3

M1: Correct amplitude and period stated
M1: Correct shape with two full cycles shown
A1: All intercepts and turning points correctly labelled

15. Solve $\sin(x + 30^\circ) = \frac{1}{2}$ for $0^\circ \leq x \leq 360^\circ$ .

Working: $x + 30^\circ = 30^\circ, 150^\circ, 390^\circ, 510^\circ$
(adding $360^\circ$ ; $x + 30^\circ$ ranges from $30^\circ$ to $390^\circ$ )

$x = 0^\circ, 120^\circ, 360^\circ$

Answer: $x = 0^\circ, 120^\circ, 360^\circ$

Marks: 3

M1: Correct principal values $x + 30^\circ = 30^\circ, 150^\circ$
M1: Consider extended range for $x + 30^\circ$
A1: All three correct values

16. Express $4\cos x - 3\sin x$ as $R\cos(x + \alpha)$ . Find maximum value and where it occurs.

Working: $R\cos(x + \alpha) = R\cos x\cos\alpha - R\sin x\sin\alpha$

Comparing: $R\cos\alpha = 4$ , $R\sin\alpha = 3$

$R = \sqrt{16 + 9} = 5$

$\tan\alpha = \frac{3}{4} \Rightarrow \alpha = \tan^{-1}\left(\frac{3}{4}\right) \approx 36.9^\circ$

Maximum value of $5\cos(x + \alpha) = 5$ when $\cos(x + \alpha) = 1$
$x + \alpha = 0^\circ$ or $360^\circ$
$x = 360^\circ - 36.9^\circ = 323.1^\circ$ (within range)

Answer: $5\cos(x + 36.9^\circ)$ ; maximum value $5$ at $x \approx 323.1^\circ$

Marks: 4

M1: Correct expansion and comparison
A1: $R = 5$ , $\alpha \approx 36.9^\circ$
A1: Maximum value $5$
A1: $x \approx 323.1^\circ$

17. Triangle $PQR$ : $PQ = 8$ cm, $QR = 11$ cm, $\angle PQR = 52^\circ$ .

(a) Length of $PR$ :

Working (Cosine Rule): $PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos\angle PQR$
$= 64 + 121 - 2(8)(11)\cos 52^\circ$
$= 185 - 176 \times 0.6157$
$= 185 - 108.36 = 76.64$
$PR = \sqrt{76.64} \approx 8.75$ cm

Answer: $PR \approx 8.75$ cm (3 s.f.)

(b) Area of triangle $PQR$ :

Working: Area $= \frac{1}{2}(PQ)(QR)\sin\angle PQR = \frac{1}{2}(8)(11)\sin 52^\circ$
$= 44 \times 0.7880 \approx 34.7$ cm $^2$

Answer: $34.7$ cm $^2$ (3 s.f.)

Marks: 4 (2+2)

(a) M1: Correct cosine rule setup; A1: Correct answer to 3 s.f.
(b) M1: Correct area formula; A1: Correct answer to 3 s.f.

18. Triangle $ABC$ : $AB = 12$ cm, $AC = 9$ cm, $\angle BAC = 65^\circ$ .

(a) Length of $BC$ :

Working (Cosine Rule): $BC^2 = 12^2 + 9^2 - 2(12)(9)\cos 65^\circ$
$= 144 + 81 - 216 \times 0.4226$
$= 225 - 91.28 = 133.72$
$BC = \sqrt{133.72} \approx 11.6$ cm

Answer: $BC \approx 11.6$ cm (3 s.f.)

(b) Length of $CD$ where $CD \perp AB$ :

Working: $CD = AC\sin\angle BAC = 9\sin 65^\circ = 9 \times 0.9063 \approx 8.16$ cm

Answer: $CD \approx 8.16$ cm (3 s.f.)

Marks: 4 (2+2)

(a) M1: Correct cosine rule; A1: Correct answer
(b) M1: Use $CD = AC\sin 65^\circ$ ; A1: Correct answer

19. Angle of elevation problem.

(a) Show $h = \frac{40}{\cot 22^\circ - \cot 35^\circ}$ .

Working: Let distance from $A$ to building base be $d$ . Then distance from $B$ to building base is $d + 40$ .

From point $A$ : $\tan 35^\circ = \frac{h}{d} \Rightarrow d = h\cot 35^\circ$
From point $B$ : $\tan 22^\circ = \frac{h}{d+40} \Rightarrow d + 40 = h\cot 22^\circ$

Subtracting: $40 = h\cot 22^\circ - h\cot 35^\circ = h(\cot 22^\circ - \cot 35^\circ)$

$h = \frac{40}{\cot 22^\circ - \cot 35^\circ}$ ✓

(b) Calculate $h$ :

Working: $\cot 22^\circ = \frac{1}{\tan 22^\circ} \approx 2.4751$
$\cot 35^\circ = \frac{1}{\tan 35^\circ} \approx 1.4281$

$h = \frac{40}{2.4751 - 1.4281} = \frac{40}{1.0470} \approx 38.2$ m

Answer: $h \approx 38.2$ m (3 s.f.)

Marks: 4 (2+2)

(a) M1: Set up two equations using $\tan$ ; A1: Correct derivation
(b) M1: Correct substitution; A1: Correct answer to 3 s.f.

20. Quadrilateral $ABCD$ : $AB = 6$ cm, $BC = 8$ cm, $CD = 5$ cm, $\angle ABC = 110^\circ$ , $\angle BCD = 70^\circ$ .

(a) Length of diagonal $AC$ :

Working (Cosine Rule in $\triangle ABC$ ): $AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos\angle ABC$
$= 36 + 64 - 2(6)(8)\cos 110^\circ$
$= 100 - 96(-0.3420)$
$= 100 + 32.83 = 132.83$
$AC = \sqrt{132.83} \approx 11.5$ cm

Answer: $AC \approx 11.5$ cm (3 s.f.)

(b) Area of quadrilateral $ABCD$ :

Working: Area $=$ Area( $\triangle ABC$ ) $+$ Area( $\triangle ACD$ )

Area( $\triangle ABC$ ) $= \frac{1}{2}(AB)(BC)\sin\angle ABC = \frac{1}{2}(6)(8)\sin 110^\circ$
$= 24 \times 0.9397 = 22.55$ cm $^2$

In $\triangle ACD$ : $AC \approx 11.53$ cm, $CD = 5$ cm, $\angle BCD = 70^\circ$

$\angle ACD = \angle BCD - \angle BCA$

First find $\angle BCA$ using sine rule in $\triangle ABC$ :
$\frac{\sin\angle BCA}{AB} = \frac{\sin\angle ABC}{AC}$
$\sin\angle BCA = \frac{6 \times \sin 110^\circ}{11.53} = \frac{6 \times 0.9397}{11.53} = \frac{5.638}{11.53} \approx 0.4890$
$\angle BCA \approx 29.3^\circ$

$\angle ACD = 70^\circ - 29.3^\circ = 40.7^\circ$

Area( $\triangle ACD$ ) $= \frac{1}{2}(AC)(CD)\sin\angle ACD$
$= \frac{1}{2}(11.53)(5)\sin 40.7^\circ$
$= 28.825 \times 0.6521 \approx 18.80$ cm $^2$

Total area $\approx 22.55 + 18.80 = 41.4$ cm $^2$

Answer: $41.4$ cm $^2$ (3 s.f.)

Marks: 4 (2+2)

(a) M1: Correct cosine rule in $\triangle ABC$ ; A1: Correct answer
(b) M1: Correct area of $\triangle ABC$ ; M1: Correct method for area of $\triangle ACD$ ; A1: Correct total area

END OF ANSWER KEY