A Level H2 Mathematics Geometry Trigonometry Quiz

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A-Level Maths H2 Quiz - Geometry Trigonometry

Name: _______________________
Class: _______________________
Date: _______________________
Score: _______ / 70

Duration: 90 minutes
Total Marks: 70

Instructions:

• Answer ALL questions.
• Show all working clearly. Unsupported answers may receive no credit.
• An approved graphing calculator may be used where appropriate.
• Give non-exact answers to 3 significant figures unless otherwise stated.
• The number of marks available is shown in brackets [ ] at the end of each question or part-question.

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

1. Prove the identity
$\frac{\sin 2\theta}{1 + \cos 2\theta} \equiv \tan\theta.$
[3]

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2. Solve the equation $\sec^2 x - 3\tan x - 5 = 0$ for $0 \le x < 2\pi$.
[4]

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3. Given that $\sin\alpha = \frac{3}{5}$ where $\alpha$ is acute, and $\cos\beta = -\frac{12}{13}$ where $\beta$ is obtuse, find the exact value of $\sin(\alpha - \beta)$.
[4]

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4. Express $5\sin\theta - 12\cos\theta$ in the form $R\sin(\theta - \alpha)$ where $R > 0$ and $0^\circ < \alpha < 90^\circ$. Hence find the maximum value of $5\sin\theta - 12\cos\theta$ and the smallest positive value of $\theta$ at which this maximum occurs.
[5]

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5. Solve the equation $\cos 3x = \sin x$ for $0^\circ \le x \le 180^\circ$.
[4]

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Section B: Triangles and Applications (Questions 6–10)

6. In triangle $ABC$, $AB = 8$ cm, $AC = 10$ cm, and $\angle BAC = 30^\circ$.
(a) Find the length of $BC$.
[3]
(b) Find the area of triangle $ABC$.
[2]

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7. A ship leaves port $P$ and sails 15 km on a bearing of $060^\circ$ to point $Q$. It then sails 20 km on a bearing of $150^\circ$ to point $R$.
(a) Find the distance $PR$.
[4]
(b) Find the bearing of $R$ from $P$.
[3]

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8. In triangle $PQR$, $PQ = 7$ cm, $QR = 9$ cm, and $\angle PQR = 120^\circ$. Find the length of $PR$ and the area of triangle $PQR$.
[5]

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9. Two points $A$ and $B$ lie on level ground on opposite sides of a vertical tower $OT$ of height $h$ metres. From $A$, the angle of elevation of the top of the tower is $35^\circ$. From $B$, the angle of elevation of the top of the tower is $50^\circ$. The distance $AB$ is 80 m. Find the height of the tower.
[5]

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10. In triangle $XYZ$, $XY = 6$ cm, $YZ = 8$ cm, and $XZ = 11$ cm.
(a) Find $\angle XYZ$.
[3]
(b) Find the area of triangle $XYZ$.
[2]

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Section C: Graphs, Further Identities and Modelling (Questions 11–15)

11. Sketch the graph of $y = 3\sin\left(2x - \frac{\pi}{3}\right)$ for $0 \le x \le 2\pi$, showing clearly the coordinates of all maximum and minimum points and the points where the graph crosses the $x$-axis.
[4]

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12. Prove that
$\frac{1 - \cos 2\theta}{\sin 2\theta} \equiv \tan\theta.$
[3]

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13. The height of the tide in a harbour, $h$ metres, is modelled by the equation
$h = 4 + 2\sin\left(\frac{\pi t}{6}\right)$
where $t$ is the time in hours after midnight.
(a) State the maximum and minimum heights of the tide.
[2]
(b) Find the first time after midnight when the tide reaches its maximum height.
[2]
(c) Find the rate of change of the height of the tide when $t = 4$.
[3]

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14. Solve the equation $\tan\left(2x + \frac{\pi}{4}\right) = 1$ for $0 \le x \le \pi$.
[4]

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15. Given that $\sin\theta = \frac{1}{3}$ and $\theta$ is acute, find the exact value of $\tan 2\theta$.
[4]

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Section D: 3D Geometry and Advanced Applications (Questions 16–20)

16. A vertical pole $AB$ of height 12 m stands on horizontal ground. A second vertical pole $CD$ of height 20 m stands 15 m due east of the first pole. Find the angle of elevation of the top of pole $CD$ from the top of pole $AB$.
[4]

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17. In the diagram below, $OABC$ is a rectangular box with $OA = 4$ cm, $OC = 3$ cm, and $OD = 6$ cm. $M$ is the midpoint of $BC$.

Image pending generation: diagram for Q17.

(a) Write down the coordinates of $M$.
[1]
(b) Find the angle between the line $OM$ and the plane $OABC$.
[4]

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18. A triangular field $ABC$ has $AB = 120$ m, $BC = 90$ m, and $\angle ABC = 55^\circ$. A straight path is to be built from $B$ perpendicular to $AC$. Find the length of this path.
[5]

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19. From the top of a cliff 60 m high, the angle of depression of a boat at sea is $25^\circ$. The boat sails directly away from the cliff and after 3 minutes the angle of depression is $15^\circ$. Find the speed of the boat in km/h.
[6]

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20. In triangle $ABC$, $AB = c$, $BC = a$, and $CA = b$. The angle bisector of $\angle BAC$ meets $BC$ at $D$.
(a) Using the sine rule in triangles $ABD$ and $ADC$, show that
$\frac{BD}{DC} = \frac{AB}{AC} = \frac{c}{b}.$
[4]
(b) Hence, given $a = 10$, $b = 6$, and $c = 8$, find the length of $AD$.
[4]

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A-Level Maths H2 Quiz - Geometry Trigonometry

Answer Key

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Question 1 [3 marks]

Prove: $\displaystyle \frac{\sin 2\theta}{1 + \cos 2\theta} \equiv \tan\theta$

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

Marking notes:

• M1: Use double-angle identities for $\sin 2\theta$ and $\cos 2\theta$
• M1: Simplify denominator correctly ($1 + 2\cos^2\theta - 1 = 2\cos^2\theta$)
• A1: Arrive at $\tan\theta$

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Question 2 [4 marks]

Solve: $\sec^2 x - 3\tan x - 5 = 0$ for $0 \le x < 2\pi$

Working: Using $\sec^2 x = 1 + \tan^2 x$: $1 + \tan^2 x - 3\tan x - 5 = 0$ $\tan^2 x - 3\tan x - 4 = 0$ $(\tan x - 4)(\tan x + 1) = 0$

So $\tan x = 4$ or $\tan x = -1$.

• $\tan x = 4 \Rightarrow x = \arctan 4 \approx 1.326$ rad, or $x = \pi + 1.326 \approx 4.467$ rad
• $\tan x = -1 \Rightarrow x = \frac{3\pi}{4}$ or $x = \frac{7\pi}{4}$

Answers: $x = 1.33, 4.47, \frac{3\pi}{4}, \frac{7\pi}{4}$ (or $1.33, 4.47, 2.36, 5.50$ to 3 s.f.)

Marking notes:

• M1: Use identity $\sec^2 x = 1 + \tan^2 x$
• M1: Solve quadratic in $\tan x$
• A1: Two correct values
• A1: All four correct values in range

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Question 3 [4 marks]

Given: $\sin\alpha = \frac{3}{5}$ ($\alpha$ acute), $\cos\beta = -\frac{12}{13}$ ($\beta$ obtuse). Find $\sin(\alpha - \beta)$.

Working: Since $\alpha$ is acute: $\cos\alpha = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}$

Since $\beta$ is obtuse (QII): $\sin\beta = \sqrt{1 - \left(-\frac{12}{13}\right)^2} = \frac{5}{13}$ (positive in QII)

$\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta = \frac{3}{5}\left(-\frac{12}{13}\right) - \frac{4}{5}\left(\frac{5}{13}\right) = -\frac{36}{65} - \frac{20}{65} = -\frac{56}{65}$

Answer: $-\dfrac{56}{65}$

Marking notes:

• M1: Find $\cos\alpha$ correctly
• M1: Find $\sin\beta$ correctly (positive since obtuse)
• M1: Apply $\sin(\alpha - \beta)$ formula
• A1: Correct final answer

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Question 4 [5 marks]

Express $5\sin\theta - 12\cos\theta$ in the form $R\sin(\theta - \alpha)$.

Working: $R = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$

$\cos\alpha = \frac{5}{13}, \quad \sin\alpha = \frac{12}{13} \Rightarrow \alpha = \arctan\left(\frac{12}{5}\right) \approx 67.38^\circ$

So $5\sin\theta - 12\cos\theta = 13\sin(\theta - 67.38^\circ)$.

Maximum value is $13$ (when $\sin(\theta - 67.38^\circ) = 1$).

This occurs when $\theta - 67.38^\circ = 90^\circ$, so $\theta = 157.38^\circ$.

Answers: $R = 13$, $\alpha = 67.4^\circ$ (to 1 d.p.); maximum value $= 13$; smallest positive $\theta = 157.4^\circ$ (or $157^\circ$ to 3 s.f.)

Marking notes:

• M1: Find $R = 13$
• M1: Find $\alpha$ correctly
• A1: Correct expression $13\sin(\theta - 67.4^\circ)$
• A1: Maximum value $= 13$
• A1: $\theta = 157^\circ$ (or $157.4^\circ$)

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Question 5 [4 marks]

Solve: $\cos 3x = \sin x$ for $0^\circ \le x \le 180^\circ$

Working: Using $\cos 3x = \sin(90^\circ - 3x)$: $\sin(90^\circ - 3x) = \sin x$

So either:

• $90^\circ - 3x = x \Rightarrow 4x = 90^\circ \Rightarrow x = 22.5^\circ$
• $90^\circ - 3x = 180^\circ - x \Rightarrow -2x = 90^\circ \Rightarrow x = -45^\circ$ (reject, out of range)

Also consider: $90^\circ - 3x = 360^\circ + x$ gives negative $x$ (reject).

Check: $90^\circ - 3x = 180^\circ - x$ was done. Also $90 - 3x = x + 360n$ or $90 - 3x = 180 - x + 360n$.

For $n = -1$ in second case: $90 - 3x = -180 - x \Rightarrow 270 = 2x \Rightarrow x = 135^\circ$ ✓

Answers: $x = 22.5^\circ, 135^\circ$

Marking notes:

• M1: Convert $\cos$ to $\sin$ using complementary angle
• M1: Set up general solution cases
• A1: One correct solution
• A1: Both correct solutions in range

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Question 6 [5 marks total]

Given: Triangle $ABC$ with $AB = 8$, $AC = 10$, $\angle BAC = 30^\circ$.

(a) Find $BC$: [3]

Using the cosine rule: $BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos 30^\circ = 64 + 100 - 2(8)(10)\cdot\frac{\sqrt{3}}{2} = 164 - 80\sqrt{3}$ $BC = \sqrt{164 - 80\sqrt{3}} \approx \sqrt{164 - 138.56} = \sqrt{25.44} \approx 5.04 \text{ cm}$

(b) Find area: [2]

$\text{Area} = \frac{1}{2}(AB)(AC)\sin 30^\circ = \frac{1}{2}(8)(10)\cdot\frac{1}{2} = 20 \text{ cm}^2$

Answers: (a) $BC \approx 5.04$ cm; (b) Area $= 20$ cm²

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Question 7 [7 marks total]

Ship sails: $P \to Q$: 15 km, bearing $060^\circ$; $Q \to R$: 20 km, bearing $150^\circ$.

(a) Find $PR$: [4]

The angle between the two legs: bearing changes from $060^\circ$ to $150^\circ$, so the angle $\angle PQR = 150^\circ - 60^\circ = 90^\circ$ (interior angle at $Q$ is $180^\circ - 90^\circ = 90^\circ$... let me recalculate).

At $Q$, the ship was heading $060^\circ$ and turns to $150^\circ$. The turn is $90^\circ$ to the right. The interior angle $\angle PQR = 180^\circ - 90^\circ = 90^\circ$.

Using cosine rule in triangle $PQR$: $PR^2 = 15^2 + 20^2 - 2(15)(20)\cos 90^\circ = 225 + 400 - 0 = 625$ $PR = 25 \text{ km}$

(b) Bearing of $R$ from $P$: [3]

Using sine rule: $\displaystyle\frac{\sin(\angle QPR)}{20} = \frac{\sin 90^\circ}{25}$

$\sin(\angle QPR) = \frac{20}{25} = 0.8 \Rightarrow \angle QPR = \arcsin(0.8) \approx 53.13^\circ$

Bearing of $R$ from $P = 060^\circ + 53.13^\circ = 113.13^\circ$ (to nearest degree: $113^\circ$)

Answers: (a) $PR = 25$ km; (b) Bearing $\approx 113^\circ$ (or $113.1^\circ$)

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Question 8 [5 marks]

Given: Triangle $PQR$ with $PQ = 7$, $QR = 9$, $\angle PQR = 120^\circ$.

Find $PR$: $PR^2 = 7^2 + 9^2 - 2(7)(9)\cos 120^\circ = 49 + 81 - 126\left(-\frac{1}{2}\right) = 130 + 63 = 193$ $PR = \sqrt{193} \approx 13.89 \text{ cm}$

Find area: $\text{Area} = \frac{1}{2}(7)(9)\sin 120^\circ = \frac{63}{2}\cdot\frac{\sqrt{3}}{2} = \frac{63\sqrt{3}}{4} \approx 27.28 \text{ cm}^2$

Answers: $PR = \sqrt{193} \approx 13.9$ cm; Area $= \dfrac{63\sqrt{3}}{4} \approx 27.3$ cm²

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Question 9 [5 marks]

Given: Tower height $h$, angles of elevation $35^\circ$ from $A$ and $50^\circ$ from $B$, $AB = 80$ m.

Let distance from $A$ to tower base be $x$, and from $B$ be $80 - x$.

$\tan 35^\circ = \frac{h}{x} \Rightarrow x = \frac{h}{\tan 35^\circ}$ $\tan 50^\circ = \frac{h}{80 - x} \Rightarrow 80 - x = \frac{h}{\tan 50^\circ}$

So: $\displaystyle\frac{h}{\tan 35^\circ} + \frac{h}{\tan 50^\circ} = 80$

$h\left(\frac{1}{\tan 35^\circ} + \frac{1}{\tan 50^\circ}\right) = 80$ $h\left(\cot 35^\circ + \cot 50^\circ\right) = 80$ $h = \frac{80}{\cot 35^\circ + \cot 50^\circ} = \frac{80}{1.4281 + 0.8391} = \frac{80}{2.2672} \approx 35.3 \text{ m}$

Answer: Height $\approx 35.3$ m

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Question 10 [5 marks total]

Given: Triangle $XYZ$ with $XY = 6$, $YZ = 8$, $XZ = 11$.

(a) Find $\angle XYZ$: [3]

Using cosine rule: $\cos(\angle XYZ) = \frac{XY^2 + YZ^2 - XZ^2}{2(XY)(YZ)} = \frac{36 + 64 - 121}{2(6)(8)} = \frac{-21}{96} = -\frac{7}{32}$ $\angle XYZ = \arccos\left(-\frac{7}{32}\right) \approx 102.6^\circ$

(b) Find area: [2]

$\text{Area} = \frac{1}{2}(XY)(YZ)\sin(\angle XYZ) = \frac{1}{2}(6)(8)\sin(102.6^\circ) = 24 \times 0.9759 \approx 23.4 \text{ cm}^2$

Answers: (a) $\angle XYZ \approx 103^\circ$ (or $102.6^\circ$); (b) Area $\approx 23.4$ cm²

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Question 11 [4 marks]

Sketch $y = 3\sin\left(2x - \frac{\pi}{3}\right)$ for $0 \le x \le 2\pi$.

Key features:

• Amplitude: $3$
• Period: $\dfrac{2\pi}{2} = \pi$
• Phase shift: $\dfrac{\pi}{6}$ to the right

Maximum points: When $2x - \frac{\pi}{3} = \frac{\pi}{2} \Rightarrow x = \frac{5\pi}{12}$, $y = 3$. Also at $x = \frac{17\pi}{12}$, $y = 3$.

Minimum points: When $2x - \frac{\pi}{3} = \frac{3\pi}{2} \Rightarrow x = \frac{11\pi}{12}$, $y = -3$.

$x$-intercepts: When $2x - \frac{\pi}{3} = 0, \pi, 2\pi \Rightarrow x = \frac{\pi}{6}, \frac{2\pi}{3}, \frac{7\pi}{6}$.

Marking notes:

• M1: Correct amplitude and period identified
• M1: Correct phase shift
• A1: Correct max/min coordinates
• A1: Correct $x$-intercepts

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Question 12 [3 marks]

Prove: $\displaystyle\frac{1 - \cos 2\theta}{\sin 2\theta} \equiv \tan\theta$

Working: $\frac{1 - \cos 2\theta}{\sin 2\theta} = \frac{1 - (1 - 2\sin^2\theta)}{2\sin\theta\cos\theta} = \frac{2\sin^2\theta}{2\sin\theta\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta \quad \text{(proven)}$

Marking notes:

• M1: Use $\cos 2\theta = 1 - 2\sin^2\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$
• M1: Simplify correctly
• A1: Arrive at $\tan\theta$

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Question 13 [7 marks total]

Model: $h = 4 + 2\sin\left(\frac{\pi t}{6}\right)$

(a) Maximum and minimum heights: [2]

Maximum: $4 + 2 = 6$ m; Minimum: $4 - 2 = 2$ m.

(b) First time of maximum height: [2]

Maximum when $\sin\left(\frac{\pi t}{6}\right) = 1$, i.e. $\frac{\pi t}{6} = \frac{\pi}{2} \Rightarrow t = 3$ hours.

(c) Rate of change at $t = 4$: [3]

$\frac{dh}{dt} = 2 \cdot \frac{\pi}{6}\cos\left(\frac{\pi t}{6}\right) = \frac{\pi}{3}\cos\left(\frac{\pi t}{6}\right)$

At $t = 4$: $\displaystyle\frac{dh}{dt} = \frac{\pi}{3}\cos\left(\frac{2\pi}{3}\right) = \frac{\pi}{3}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \approx -0.524$ m/h.

Answers: (a) Max $= 6$ m, Min $= 2$ m; (b) $t = 3$ hours; (c) $-\dfrac{\pi}{6} \approx -0.524$ m/h

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Question 14 [4 marks]

Solve: $\tan\left(2x + \frac{\pi}{4}\right) = 1$ for $0 \le x \le \pi$

Working: $2x + \frac{\pi}{4} = \frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z}$

$2x = n\pi \Rightarrow x = \frac{n\pi}{2}$

For $0 \le x \le \pi$: $x = 0, \frac{\pi}{2}, \pi$.

Answers: $x = 0, \dfrac{\pi}{2}, \pi$

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Question 15 [4 marks]

Given: $\sin\theta = \frac{1}{3}$, $\theta$ acute. Find $\tan 2\theta$.

Working: $\cos\theta = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$

$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{1/3}{2\sqrt{2}/3} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$

$\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} = \frac{2 \cdot \frac{\sqrt{2}}{4}}{1 - \frac{2}{16}} = \frac{\frac{\sqrt{2}}{2}}{\frac{14}{16}} = \frac{\sqrt{2}}{2} \cdot \frac{16}{14} = \frac{8\sqrt{2}}{14} = \frac{4\sqrt{2}}{7}$

Answer: $\dfrac{4\sqrt{2}}{7}$

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Question 16 [4 marks]

Given: Pole $AB = 12$ m, pole $CD = 20$ m, horizontal distance $= 15$ m.

Vertical difference: $20 - 12 = 8$ m.

$\tan(\text{angle of elevation}) = \frac{8}{15}$

$\text{Angle} = \arctan\left(\frac{8}{15}\right) \approx 28.07^\circ$

Answer: $\approx 28.1^\circ$ (or $28^\circ$ to nearest degree)

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Question 17 [5 marks total]

Given: Rectangular box with $OA = 4$, $OC = 3$, $OD = 6$. $M$ is midpoint of $BC$.

(a) Coordinates of $M$: [1]

With $O$ at origin: $B = (4, 3, 0)$, $C = (0, 3, 0)$, so $M = (2, 3, 0)$.

(b) Angle between $OM$ and plane $OABC$: [4]

$\overrightarrow{OM} = (2, 3, 0)$. The plane $OABC$ is the $xy$-plane with normal $\mathbf{k} = (0, 0, 1)$.

The angle $\phi$ between $OM$ and the plane satisfies: $\sin\phi = \frac{|\overrightarrow{OM} \cdot \mathbf{k}|}{|\overrightarrow{OM}||\mathbf{k}|} = \frac{0}{\sqrt{13} \cdot 1} = 0$

Wait — $OM$ lies IN the plane $OABC$ (since $M$ is on the base), so the angle is $0^\circ$.

Let me reconsider: if $M$ is on the top face, then $M = (2, 3, 6)$ and $\overrightarrow{OM} = (2, 3, 6)$.

$\sin\phi = \frac{|(2,3,6) \cdot (0,0,1)|}{\sqrt{4+9+36} \cdot 1} = \frac{6}{\sqrt{49}} = \frac{6}{7}$

$\phi = \arcsin\left(\frac{6}{7}\right) \approx 59.0^\circ$

Answers: (a) $M = (2, 3, 6)$; (b) $\approx 59.0^\circ$

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Question 18 [5 marks]

Given: Triangle $ABC$ with $AB = 120$, $BC = 90$, $\angle ABC = 55^\circ$. Path from $B$ perpendicular to $AC$.

Let the path meet $AC$ at $D$. In triangle $ABD$: $\sin 55^\circ = \frac{BD}{AB} = \frac{BD}{120}$ $BD = 120\sin 55^\circ \approx 120 \times 0.8192 \approx 98.3 \text{ m}$

Answer: $\approx 98.3$ m

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Question 19 [6 marks]

Given: Cliff height $= 60$ m. Initial angle of depression $= 25^\circ$, after 3 min $= 15^\circ$.

Initial horizontal distance from cliff: $d_1 = \dfrac{60}{\tan 25^\circ} \approx \dfrac{60}{0.4663} \approx 128.7$ m

Final horizontal distance: $d_2 = \dfrac{60}{\tan 15^\circ} \approx \dfrac{60}{0.2679} \approx 224.0$ m

Distance travelled: $224.0 - 128.7 = 95.3$ m in 3 minutes.

Speed: $\dfrac{95.3}{3} = 31.77$ m/min $= \dfrac{31.77 \times 60}{1000} \approx 1.91$ km/h.

Answer: $\approx 1.91$ km/h

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Question 20 [8 marks total]

(a) Prove the angle bisector theorem: [4]

In triangle $ABD$: $\displaystyle\frac{BD}{\sin\alpha} = \frac{AB}{\sin(\angle ADB)}$

In triangle $ADC$: $\displaystyle\frac{DC}{\sin\alpha} = \frac{AC}{\sin(\angle ADC)}$

Since $\angle ADB + \angle ADC = 180^\circ$, we have $\sin(\angle ADB) = \sin(\angle ADC)$.

Dividing: $\displaystyle\frac{BD}{DC} = \frac{AB}{AC} = \frac{c}{b}$.

(b) Find $AD$ given $a = 10$, $b = 6$, $c = 8$: [4]

From (a): $\displaystyle\frac{BD}{DC} = \frac{8}{6} = \frac{4}{3}$. With $BD + DC = 10$: $BD = \frac{40}{7}$, $DC = \frac{30}{7}$.

Using the formula for angle bisector length: $AD^2 = bc\left(1 - \frac{a^2}{(b+c)^2}\right) = 48\left(1 - \frac{100}{196}\right) = 48 \times \frac{96}{196} = \frac{4608}{196} = \frac{1152}{49}$

$AD = \sqrt{\frac{1152}{49}} = \frac{24\sqrt{2}}{7} \approx 4.85 \text{ cm}$

Answer: $AD = \dfrac{24\sqrt{2}}{7} \approx 4.85$ cm