A Level H2 Mathematics Geometry Trigonometry Quiz

A-Level Maths H2 Quiz - Geometry Trigonometry (Answer Key)

1. Solve $\sin(2x) = \frac{\sqrt{3}}{2}$ for $0 \le x \le 2\pi$.

• Basic angle for $\sin \alpha = \frac{\sqrt{3}}{2}$ is $\frac{\pi}{3}$.
• $2x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{7\pi}{3}, \frac{8\pi}{3}$ (within range $0 \le 2x \le 4\pi$).
• $x = \frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}$.
• Answer: $x = \frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}$ [2]

2. Given $\cos \theta = -\frac{3}{5}$, $\pi < \theta < \frac{3\pi}{2}$ (3rd Quadrant).

• In 3rd Quadrant, $\sin \theta < 0$ and $\tan \theta > 0$.
• Using $\sin^2 \theta + \cos^2 \theta = 1$: $\sin^2 \theta = 1 - (-\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}$.
• $\sin \theta = -\frac{4}{5}$.
• $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-4/5}{-3/5} = \frac{4}{3}$.
• Answer: $\frac{4}{3}$ [2]

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

• Substitute $\cos^2 x = 1 - \sin^2 x$: $2(1 - \sin^2 x) - \sin x - 1 = 0$ $2 - 2\sin^2 x - \sin x - 1 = 0$ $2\sin^2 x + \sin x - 1 = 0$
• Factorize: $(2\sin x - 1)(\sin x + 1) = 0$.
• $\sin x = \frac{1}{2}$ or $\sin x = -1$.
• For $\sin x = \frac{1}{2}$: $x = 30^\circ, 150^\circ$.
• For $\sin x = -1$: $x = 270^\circ$.
• Answer: $30^\circ, 150^\circ, 270^\circ$ [3]

4. Prove $\frac{1 - \cos 2A}{\sin 2A} = \tan A$.

• LHS: Use double angle formulas $\cos 2A = 1 - 2\sin^2 A$ and $\sin 2A = 2\sin A \cos A$.
• Numerator: $1 - (1 - 2\sin^2 A) = 2\sin^2 A$.
• Denominator: $2\sin A \cos A$.
• LHS $= \frac{2\sin^2 A}{2\sin A \cos A} = \frac{\sin A}{\cos A} = \tan A =$ RHS.
• Answer: Shown [2]

5. Exact value of $\sin(75^\circ)$.

• $\sin(75^\circ) = \sin(45^\circ + 30^\circ)$.
• Formula: $\sin(A+B) = \sin A \cos B + \cos A \sin B$.
• $= \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$
• $= \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)$
• $= \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}}$.
• Rationalize denominator: $\frac{(\sqrt{3}+1)\sqrt{2}}{4} = \frac{\sqrt{6}+\sqrt{2}}{4}$.
• Answer: $\frac{\sqrt{6}+\sqrt{2}}{4}$ [2]

6. Express $3\cos x - 4\sin x$ as $R\cos(x + \alpha)$.

• $R = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = 5$.
• $3\cos x - 4\sin x = 5(\frac{3}{5}\cos x - \frac{4}{5}\sin x)$.
• Compare with $R(\cos x \cos \alpha - \sin x \sin \alpha)$.
• $\cos \alpha = \frac{3}{5}, \sin \alpha = \frac{4}{5}$.
• $\tan \alpha = \frac{4}{3} \implies \alpha = \arctan(\frac{4}{3}) \approx 0.927$ rad.
• Answer: $R=5, \alpha \approx 0.927$ [3]

7. Solve $3\cos x - 4\sin x = 2$.

• $5\cos(x + 0.927) = 2 \implies \cos(x + 0.927) = 0.4$.
• Basic angle: $\arccos(0.4) \approx 1.159$.
• $x + 0.927 = 1.159$ or $2\pi - 1.159$.
• $x_1 = 1.159 - 0.927 = 0.232$.
• $x_2 = (2\pi - 1.159) - 0.927 = 5.124 - 0.927 = 4.197$.
• Answer: $x \approx 0.232, 4.20$ (3 s.f.) [3]

8. Max value of $5\sin x + 12\cos x + 4$.

• Let $y = 5\sin x + 12\cos x$. $R = \sqrt{5^2+12^2} = 13$.
• Max value of $5\sin x + 12\cos x$ is $13$.
• Max value of expression $= 13 + 4 = 17$.
• Occurs when $\sin(x+\alpha) = 1$.
• Form: $13\sin(x+\alpha)$. $\cos \alpha = 5/13, \sin \alpha = 12/13$.
• $\alpha = \arcsin(12/13) \approx 1.176$ rad.
• $x + 1.176 = \frac{\pi}{2} \implies x = \frac{\pi}{2} - 1.176 \approx 0.395$.
• Answer: Max Value = 17, $x \approx 0.395$ rad [3]

9. Sketch $y = 2\sin(2x)$ for $0 \le x \le \pi$.

• Period: $\frac{2\pi}{2} = \pi$. Amplitude: 2.
• Intercepts: $(0,0), (\frac{\pi}{2}, 0), (\pi, 0)$.
• Max at $x = \frac{\pi}{4}, y=2$. Min at $x = \frac{3\pi}{4}, y=-2$.
• Answer: Sine wave starting at origin, peak at $(\frac{\pi}{4}, 2)$, crossing axis at $\frac{\pi}{2}$, trough at $(\frac{3\pi}{4}, -2)$, ending at $(\pi, 0)$. [3]

10. Graph $y = a \cos(bx) + c$. Max 5, Min -1, Period $\pi$.

• Amplitude $a = \frac{\text{Max} - \text{Min}}{2} = \frac{5 - (-1)}{2} = 3$.
• Vertical shift $c = \frac{\text{Max} + \text{Min}}{2} = \frac{5 + (-1)}{2} = 2$.
• Period $= \frac{2\pi}{b} = \pi \implies b = 2$.
• Answer: $a=3, b=2, c=2$ [3]

11. Triangle $ABC$, $AB=10, AC=8, \angle B = 30^\circ$. Find $\angle C$.

• Sine Rule: $\frac{\sin C}{10} = \frac{\sin 30^\circ}{8}$.
• $\sin C = \frac{10 \sin 30^\circ}{8} = \frac{10(0.5)}{8} = \frac{5}{8} = 0.625$.
• $C_1 = \arcsin(0.625) \approx 38.7^\circ$.
• $C_2 = 180^\circ - 38.7^\circ = 141.3^\circ$.
• Check validity: $30 + 141.3 < 180$, so both are valid.
• Answer: $38.7^\circ$ and $141.3^\circ$ [3]

12. Area using larger $\angle C = 141.3^\circ$.

• $\angle A = 180 - 30 - 141.3 = 8.7^\circ$.
• Area $= \frac{1}{2} bc \sin A = \frac{1}{2}(8)(10)\sin(8.7^\circ)$.
• Area $= 40 \sin(8.7^\circ) \approx 6.06$ cm$^2$.
• Answer: $6.06$ cm$^2$ [2]

13. Triangle $PQR$, $PQ=7, QR=9, \angle Q = 120^\circ$. Find $PR$.

• Cosine Rule: $PR^2 = 7^2 + 9^2 - 2(7)(9)\cos(120^\circ)$.
• $PR^2 = 49 + 81 - 126(-0.5)$.
• $PR^2 = 130 + 63 = 193$.
• $PR = \sqrt{193} \approx 13.9$.
• Answer: $13.9$ [2]

14. Tower height $h$. $\angle B = 45^\circ \implies AB = h$. $\angle C = 30^\circ \implies AC = h\sqrt{3}$ (since $\tan 30 = h/AC \implies AC = h/\tan 30 = h\sqrt{3}$).

• In $\triangle ABC$, by Cosine Rule on side $BC$: $BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(60^\circ)$. $50^2 = h^2 + (h\sqrt{3})^2 - 2(h)(h\sqrt{3})(\frac{1}{2})$. $2500 = h^2 + 3h^2 - h^2\sqrt{3}$. $2500 = 4h^2 - h^2\sqrt{3}$. $2500 = h^2(4 - \sqrt{3})$. Correction in Question Prompt logic check: The prompt asks to show $h^2(3-\sqrt{3})=2500$. Let's re-read carefully. Ah, standard problem usually has specific angles. Let's re-evaluate the geometry. If $\angle BAC = 60$, $AB=h$, $AC=h\sqrt{3}$. $BC^2 = h^2 + 3h^2 - 2h(h\sqrt{3})(0.5) = 4h^2 - h^2\sqrt{3} = h^2(4-\sqrt{3})$. The question statement in the quiz says "show that $h^2(3-\sqrt{3})=2500$". This implies a discrepancy in the standard setup or my derivation. Let's check the angles again. Elevation from B is 45 ($AB=h$). Elevation from C is 30 ($AC=h\sqrt{3}$). Is it possible the angle at A is different? No, given as 60. Let's check the target expression: $3-\sqrt{3}$. If the question intended $AC = h \cot 30 = h\sqrt{3}$ and $AB = h \cot 45 = h$. Maybe the angle given was different? Or the target expression in the prompt text was a typo for $4-\sqrt{3}$? Self-Correction for Answer Key: I will provide the derivation for $h^2(4-\sqrt{3})=2500$ as that is mathematically correct for the stated parameters. If the student follows the prompt's "Show that", they might be stuck. I will note the likely typo in the question design or assume the angle at A was such that it yields $3-\sqrt{3}$. Actually, if $\angle BAC = 90^\circ$, $BC^2 = h^2 + 3h^2 = 4h^2$. If $\angle BAC = 30^\circ$? $BC^2 = h^2 + 3h^2 - 2h(h\sqrt{3})(\frac{\sqrt{3}}{2}) = 4h^2 - 3h^2 = h^2$. Let's stick to the calculation derived from the text provided: $h^2(4-\sqrt{3}) = 2500$. Note: In a real exam, if the "Show that" doesn't match, students check their work. Here, I will provide the correct mathematical result for the given numbers. Answer: Derived $h^2(4-\sqrt{3}) = 2500$. [4]

15. Find $h$.

• $h = \sqrt{\frac{2500}{4-\sqrt{3}}} \approx \sqrt{\frac{2500}{2.268}} \approx \sqrt{1102.3} \approx 33.2$.
• Answer: $33.2$ m [1]

16. $\tan A = 1/2, \tan B = 1/3$.

• $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{1/2 + 1/3}{1 - (1/2)(1/3)} = \frac{5/6}{1 - 1/6} = \frac{5/6}{5/6} = 1$.
• Since $A, B$ acute, $0 < A+B < \pi$.
• $\tan(A+B) = 1 \implies A+B = \frac{\pi}{4}$.
• Answer: $1, \frac{\pi}{4}$ [3]

17. Solve $\sin x + \sqrt{3}\cos x = 1$.

• $R = \sqrt{1^2 + (\sqrt{3})^2} = 2$.
• $2(\frac{1}{2}\sin x + \frac{\sqrt{3}}{2}\cos x) = 1$.
• Form $R\sin(x+\alpha)$: $\cos \alpha = 1/2, \sin \alpha = \sqrt{3}/2 \implies \alpha = \frac{\pi}{3}$.
• $2\sin(x + \frac{\pi}{3}) = 1 \implies \sin(x + \frac{\pi}{3}) = 0.5$.
• $x + \frac{\pi}{3} = \frac{\pi}{6}, \frac{5\pi}{6}$.
• $x = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6}$ (Reject, out of range). Add $2\pi \implies \frac{11\pi}{6}$.
• $x = \frac{5\pi}{6} - \frac{\pi}{3} = \frac{3\pi}{6} = \frac{\pi}{2}$.
• Answer: $x = \frac{\pi}{2}, \frac{11\pi}{6}$ [4]

18. Small angle approximations.

• $\sin \theta \approx \theta$, $\cos \theta \approx 1 - \frac{\theta^2}{2}$.
• LHS $= \frac{1 - (1 - \theta^2/2)}{\theta} = \frac{\theta^2/2}{\theta} = \frac{\theta}{2}$.
• Answer: Shown [3]

19. Prove $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$.

• $\cos 3\theta = \cos(2\theta + \theta) = \cos 2\theta \cos \theta - \sin 2\theta \sin \theta$.
• $= (2\cos^2 \theta - 1)\cos \theta - (2\sin \theta \cos \theta)\sin \theta$.
• $= 2\cos^3 \theta - \cos \theta - 2\sin^2 \theta \cos \theta$.
• Substitute $\sin^2 \theta = 1 - \cos^2 \theta$:
• $= 2\cos^3 \theta - \cos \theta - 2(1 - \cos^2 \theta)\cos \theta$.
• $= 2\cos^3 \theta - \cos \theta - 2\cos \theta + 2\cos^3 \theta$.
• $= 4\cos^3 \theta - 3\cos \theta$.
• Answer: Shown [3]

20. Solve $8\cos^3 \theta - 6\cos \theta - 1 = 0$.

• Factor out 2: $2(4\cos^3 \theta - 3\cos \theta) = 1$.
• Using Q19: $2\cos 3\theta = 1 \implies \cos 3\theta = 0.5$.
• $3\theta = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}$ (Range for $\theta \in [0, \pi] \implies 3\theta \in [0, 3\pi]$).
• $\theta = \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}$.
• Answer: $\frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}$ [3]