O Level Additional Mathematics Geometry Trigonometry Quiz

Answer Key - O-Level Additional Mathematics Quiz (Geometry Trigonometry)

1. $2(1 - \sin^2 \theta) + 3\sin \theta = 3 \implies 2\sin^2 \theta - 3\sin \theta + 1 = 0$. $(2\sin \theta - 1)(\sin \theta - 1) = 0$. $\sin \theta = 0.5 \implies \theta = 30^\circ, 150^\circ$. $\sin \theta = 1 \implies \theta = 90^\circ$. Ans: $30^\circ, 90^\circ, 150^\circ$

2. LHS $= \frac{1 - (1 - 2\sin^2 A)}{\sin 2A} = \frac{2\sin^2 A}{2\sin A \cos A} = \frac{\sin A}{\cos A} = \tan A$. (Proven)

3. $\tan \alpha = 3/4$ in 3rd quadrant. $\cos \alpha$ is negative. $\sec^2 \alpha = 1 + \tan^2 \alpha = 1 + 9/16 = 25/16$. $\cos^2 \alpha = 16/25 \implies \cos \alpha = -4/5$. Ans: $-0.8$

4. $2\theta - 30^\circ = 60^\circ, 240^\circ, 420^\circ, 660^\circ \dots$ $2\theta = 90^\circ, 270^\circ, 450^\circ, 690^\circ \dots$ $\theta = 45^\circ, 135^\circ$. Ans: $45^\circ, 135^\circ$

5. $R = \sqrt{5^2 + (-12)^2} = 13$. $\tan \alpha = 12/5 \implies \alpha = 67.4^\circ$. Ans: $13\cos(\theta + 67.4^\circ)$

6. $R = 5, \alpha = \tan^{-1}(3/4) = 36.9^\circ$. $5\sin(\theta + 36.9^\circ) = 2 \implies \sin(\theta + 36.9^\circ) = 0.4$. $\theta + 36.9^\circ = 23.6^\circ$ (negative $\theta$) or $156.4^\circ$. $\theta = 156.4^\circ - 36.9^\circ = 119.5^\circ$. Ans: $119.5^\circ$

7. $2\sin x \cos x = \cos x \implies \cos x(2\sin x - 1) = 0$. $\cos x = 0 \implies x = \pi/2, 3\pi/2$. $\sin x = 1/2 \implies x = \pi/6, 5\pi/6$. Ans: $\pi/6, \pi/2, 5\pi/6, 3\pi/2$

8. $\sec^2 A - \tan^2 A = \frac{1}{\cos^2 A} - \frac{\sin^2 A}{\cos^2 A} = \frac{1 - \sin^2 A}{\cos^2 A} = \frac{\cos^2 A}{\cos^2 A} = 1$. (Proven)

9. Amplitude $= 3$, Period $= \pi$. Graph: Sine wave starting at $(0,0)$, peak at $(\pi/4, 3)$, crossing at $(\pi/2, 0)$, trough at $(3\pi/4, -3)$, ending at $(\pi, 0)$.

10. $(2\sin \theta + 1)(\sin \theta - 1) = 0$. $\sin \theta = -0.5 \implies \theta = 210^\circ, 330^\circ$. $\sin \theta = 1 \implies \theta = 90^\circ$. Ans: $90^\circ, 210^\circ, 330^\circ$

11. $2x + 1 = x^2 - 2 \implies x^2 - 2x - 3 = 0 \implies (x-3)(x+1) = 0$. $x = 3 \implies y = 7$; $x = -1 \implies y = -1$. Ans: $(3, 7)$ and $(-1, -1)$

12. $(x-3)^2 - 9 + (y+4)^2 - 16 + 9 = 0 \implies (x-3)^2 + (y+4)^2 = 16$. Ans: Centre $(3, -4)$, Radius $= 4$

13. Midpoint $M = ((-1+5)/2, (4+2)/2) = (2, 3)$. $r^2 = (2 - (-1))^2 + (3-4)^2 = 3^2 + (-1)^2 = 10$. Ans: $(x-2)^2 + (y-3)^2 = 10$

14. Centre $C(2, -1)$. Gradient $C$ to $(5, 3) = \frac{3 - (-1)}{5 - 2} = \frac{4}{3}$. Gradient of tangent $L = -3/4$. $y - 3 = -3/4(x - 5) \implies 4y - 12 = -3x + 15 \implies 3x + 4y = 27$. Ans: $3x + 4y = 27$

15. Midpoint $M = (4, 1)$. Gradient $PQ = \frac{5 - (-3)}{6 - 2} = \frac{8}{4} = 2$. Perpendicular gradient $= -1/2$. $y - 1 = -1/2(x - 4) \implies 2y - 2 = -x + 4 \implies x + 2y = 6$. Ans: $x + 2y = 6$

16. $18 = a(3^2) \implies 9a = 18 \implies a = 2$. Curve: $y = 2x^2$. $\frac{dy}{dx} = 4x$. At $x=3$, gradient $= 12$. $y - 18 = 12(x - 3) \implies y = 12x - 18$. Ans: $a=2, y = 12x - 18$

17. $(x+2)^2 - 4 + (y-3)^2 - 9 - 12 = 0 \implies (x+2)^2 + (y-3)^2 = 25$. $r = \sqrt{25} = 5$. (Shown)

18. $x = \frac{3(1) + 2(6)}{5} = \frac{15}{5} = 3$; $y = \frac{3(2) + 2(7)}{5} = \frac{20}{5} = 4$. Ans: $(3, 4)$

19. $x^2 + 2x + 1 = mx + 4 \implies x^2 + (2-m)x - 3 = 0$. For tangency, $\Delta = 0 \implies (2-m)^2 - 4(1)(-3) = 0$. $(2-m)^2 = -12$. No real values of $m$. Correction check: If $y = mx + 4$ is tangent to $y = x^2 + 2x + 1$, the discriminant must be 0. Since $(2-m)^2 + 12 = 0$ has no real solution, the line cannot be tangent. (Note to teacher: In a real exam, the constant 4 would be adjusted to e.g. -2 to allow solutions).

20. Intersection of $y=x$ and $y=-x+4$: $x = -x+4 \implies 2x=4 \implies x=2, y=2$. Vertices: $(0,0), (4,0), (2,2)$. Area $= 1/2 \times \text{base} \times \text{height} = 1/2 \times 4 \times 2 = 4$. Ans: 4 sq units