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

TuitionGoWhere Practice Paper - Elementary Mathematics Secondary 3

Answer Key & Marking Scheme (Version 2)

Subject: Elementary Mathematics
Level: Secondary 3
Assessment: SA2 Practice Paper

Section A

1.
(a) Using Pythagoras' Theorem:
$AC^2 = AB^2 + BC^2$
$AC^2 = 12^2 + 5^2 = 144 + 25 = 169$
$AC = \sqrt{169} = 13$
Answer: 13 cm [1]

(b) $\sin(\angle BAC) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC}$
$\sin(\angle BAC) = \frac{5}{13}$
Answer: $\frac{5}{13}$ [2]
(1 mark for correct ratio setup, 1 mark for final simplified fraction)

2.
Using the quadratic formula for $2x^2 - 7x - 4 = 0$:
$a=2, b=-7, c=-4$
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$x = \frac{7 \pm \sqrt{(-7)^2 - 4(2)(-4)}}{2(2)}$
$x = \frac{7 \pm \sqrt{49 + 32}}{4}$
$x = \frac{7 \pm \sqrt{81}}{4}$
$x = \frac{7 \pm 9}{4}$
$x_1 = \frac{16}{4} = 4$
$x_2 = \frac{-2}{4} = -0.5$
Answer: $x = 4.00$ or $x = -0.50$ [3]
(1 mark for substitution, 1 mark for correct roots, 1 mark for correct rounding/format)

3.
Let $M$ be the midpoint of $AB$. $AM = MB = 4$ cm.
The projection of $E$ onto the base is $A$. However, the angle is between $EM$ and the base.
We need the right-angled triangle formed by the height $EA$ and the distance $AM$ on the base? No, $E$ is above $A$. The line is $EM$. The projection of $E$ on the base is $A$. So the triangle is $\triangle EAM$.
$\angle EAM = 90^\circ$.
Height $EA = 10$ cm.
Base $AM = 4$ cm.
Let $\theta$ be the angle between $EM$ and the base ($AM$).
$\tan \theta = \frac{EA}{AM} = \frac{10}{4} = 2.5$
$\theta = \tan^{-1}(2.5) \approx 68.198^\circ$
Answer: $68.2^\circ$ [3]
(1 mark for identifying correct triangle/dimensions, 1 mark for trig ratio, 1 mark for answer)

4.
$3x^2 - 12y^2$
Factor out common factor 3:
$= 3(x^2 - 4y^2)$
Recognize difference of two squares:
$= 3(x - 2y)(x + 2y)$
Answer: $3(x - 2y)(x + 2y)$ [3]
(1 mark for factor 3, 1 mark for difference of squares structure, 1 mark for final answer)

5.
$\sin \theta = 0.6 = \frac{3}{5}$.
Since $90^\circ < \theta < 180^\circ$ (2nd quadrant), $\cos \theta$ is negative.
Using $\sin^2 \theta + \cos^2 \theta = 1$:
$(0.6)^2 + \cos^2 \theta = 1$
$0.36 + \cos^2 \theta = 1$
$\cos^2 \theta = 0.64$
$\cos \theta = -\sqrt{0.64}$ (negative because 2nd quadrant)
$\cos \theta = -0.8$ or $-\frac{4}{5}$
Answer: $-0.8$ or $-\frac{4}{5}$ [3]
(1 mark for identity/substitution, 1 mark for recognizing sign, 1 mark for answer)

6.
(a) Gradient $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 5}{8 - 2} = \frac{-4}{6} = -\frac{2}{3}$
Answer: $-\frac{2}{3}$ [1]

(b) Midpoint of $AB = (\frac{2+8}{2}, \frac{5+1}{2}) = (5, 3)$.
Gradient of perpendicular bisector $m_{\perp} = -\frac{1}{m} = \frac{3}{2} = 1.5$.
Equation: $y - y_1 = m(x - x_1)$
$y - 3 = 1.5(x - 5)$
$y = 1.5x - 7.5 + 3$
$y = 1.5x - 4.5$
Answer: $y = 1.5x - 4.5$ [2]
(1 mark for correct gradient/midpoint, 1 mark for final equation)

7.
Area $= \frac{1}{2} ab \sin C$
Area $= \frac{1}{2} (10)(14) \sin 60^\circ$
Area $= 70 \times \frac{\sqrt{3}}{2} = 35\sqrt{3}$
Area $\approx 60.621...$
Answer: $60.6$ cm$^2$ [3]
(1 mark for formula, 1 mark for substitution, 1 mark for answer)

8.
Split into two inequalities:

1. $3x - 5 < 2x + 4 \Rightarrow x < 9$
2. $2x + 4 \le 10 \Rightarrow 2x \le 6 \Rightarrow x \le 3$
   Intersection of $x < 9$ and $x \le 3$ is $x \le 3$.
   Answer: $x \le 3$ [2]
   Number line: Solid dot at 3, arrow to the left. [1]

9.
(a) Arc Length $= \frac{\theta}{360} \times 2\pi r$
$= \frac{120}{360} \times 2 \times \pi \times 9$
$= \frac{1}{3} \times 18\pi = 6\pi$
$\approx 18.849...$
Answer: $18.8$ cm [2]

(b) Area $= \frac{\theta}{360} \times \pi r^2$
$= \frac{1}{3} \times \pi \times 81 = 27\pi$
$\approx 84.823...$
Answer: $84.8$ cm$^2$ [1]

10.
Draw North lines at A, B, C.
Bearing $A \to B = 050^\circ$.
Interior angle at B (from North line at B to BA): Since North lines are parallel, co-interior angles sum to 180? No, alternate angles.
Angle of North at B to line BA is $180 + 50 = 230$? No.
Let's use geometry.
Line AB makes $50^\circ$ with North at A.
At B, the bearing of A is $050 + 180 = 230^\circ$.
Bearing of C from B is $140^\circ$.
Angle $\angle ABC = 230^\circ - 140^\circ = 90^\circ$.
So $\triangle ABC$ is right-angled isosceles at B.
$\angle BCA = 45^\circ$.
Bearing of B from C: Bearing $B \to C$ is $140^\circ$. Bearing $C \to B$ is $140 + 180 = 320^\circ$.
We need Bearing $C \to A$.
In $\triangle ABC$, angle at C is $45^\circ$.
Line CB is at bearing $320^\circ$ from C.
Line CA is to the "left" of CB?
Let's check coordinates.
$A=(0,0)$. $B = (d \sin 50, d \cos 50)$.
$C$ is reached from B by bearing 140.
Vector $BC$ direction is $140^\circ$.
Vector $BA$ direction is $230^\circ$.
Angle $ABC = 230 - 140 = 90^\circ$.
Triangle is isosceles right-angled.
Angle $BCA = 45^\circ$.
Bearing $C \to B$ is $320^\circ$.
To get to A, we turn $45^\circ$ clockwise or anti-clockwise?
A is "behind" B relative to C?
Let's visualize. A is SW of B? No, A is origin. B is NE. C is SE of B.
So C is East of A.
Bearing $C \to B$ is $320^\circ$ (NW).
A is to the West of C?
Angle $BCA = 45^\circ$.
Since A is to the left of vector CB (looking from C to B), we subtract 45?
Bearing $C \to A = 320^\circ - 45^\circ = 275^\circ$.
Answer: $275^\circ$ [3]
(1 mark for finding angle ABC=90, 1 mark for geometry of isosceles, 1 mark for final bearing calculation)

Section B

11.
(a) In $\triangle TBP$, $\angle TBP = 45^\circ$, $\angle TPB = 90^\circ$.
$\tan 45^\circ = \frac{TP}{BP} = 1 \Rightarrow BP = h$.
Answer: $h$ [1]

(b) In $\triangle TAP$, $\angle TAP = 30^\circ$.
$\tan 30^\circ = \frac{TP}{AP} = \frac{1}{\sqrt{3}}$.
$AP = h\sqrt{3}$.
Answer: $h\sqrt{3}$ [1]

(c) $AP - BP = AB = 20$.
$h\sqrt{3} - h = 20$
$h(\sqrt{3} - 1) = 20$
$h = \frac{20}{\sqrt{3} - 1}$
$h = \frac{20}{1.73205 - 1} = \frac{20}{0.73205} \approx 27.32$
Answer: $27.3$ m [4]
(1 mark for setting up equation, 1 mark for algebraic manipulation, 1 mark for correct value, 1 mark for rounding)

12.
(a) Angles in the same segment subtended by arc AD.
$\angle ACD = \angle ABD = 35^\circ$.
Answer: $35^\circ$ [1]

(b) In $\triangle ABX$:
$\angle BAX = 40^\circ$ (given as $\angle BAC$)
$\angle ABX = 35^\circ$ (given as $\angle ABD$)
$\angle AXB = 180^\circ - (40^\circ + 35^\circ) = 180^\circ - 75^\circ = 105^\circ$.
Answer: $105^\circ$ [2]

(c) $AD = DC \Rightarrow$ Arc AD = Arc DC.
$\angle DAC$ subtends Arc DC.
$\angle DCA$ subtends Arc AD.
So $\angle DAC = \angle DCA$.
We know $\angle ACD = 35^\circ$ from (a).
Wait, $\angle ACD$ is the whole angle C in $\triangle ACD$? No, $\angle ACD$ is angle subtended by AD.
$\angle DAC$ is subtended by DC.
Since chords AD=DC, angles subtended at circumference are equal.
$\angle DAC = \angle ABD$? No.
$\angle DAC$ subtends arc DC. $\angle DBC$ subtends arc DC.
$\angle ACD$ subtends arc AD. $\angle ABD$ subtends arc AD.
Given $AD=DC$, $\triangle ADC$ is isosceles.
Also $\angle DAC = \angle DCA$.
We found $\angle ACD = 35^\circ$ in (a).
Therefore $\angle DAC = 35^\circ$.
Answer: $35^\circ$ [3]
(1 mark for identifying isosceles/equal arcs, 1 mark for linking to previous angle, 1 mark for answer)

13.
(a) Completing the square:
$x^2 - 6x + 5$
$= (x - 3)^2 - 3^2 + 5$
$= (x - 3)^2 - 9 + 5$
$= (x - 3)^2 - 4$
Answer: $(x - 3)^2 - 4$ [2]

(b) Minimum point is at vertex $(3, -4)$.
Answer: $(3, -4)$ [1]

(c) Sketch:

• Parabola opening upwards.
• Vertex at $(3, -4)$.
• y-intercept: Let $x=0, y=5$. Point $(0,5)$.
• x-intercepts: $(x-3)^2=4 \Rightarrow x-3=\pm 2 \Rightarrow x=1, 5$. Points $(1,0), (5,0)$.
• Labels correct.
  [3]
  (1 mark for shape/vertex, 1 mark for intercepts, 1 mark for labels/accuracy)

14.
(a) Cosine Rule: $b^2 = a^2 + c^2 - 2ac \cos B$
$15^2 = 14^2 + 13^2 - 2(14)(13) \cos B$
$225 = 196 + 169 - 364 \cos B$
$225 = 365 - 364 \cos B$
$364 \cos B = 365 - 225 = 140$
$\cos B = \frac{140}{364} = \frac{35}{91} = \frac{5}{13}$
$B = \cos^{-1}(\frac{5}{13}) \approx 67.38^\circ$
Answer: $67.4^\circ$ [3]

(b) Area $= \frac{1}{2} ac \sin B$
$= \frac{1}{2} (14)(13) \sin(67.38^\circ)$
$= 91 \times 0.923...$
Alternatively, $\sin B = \sqrt{1 - (5/13)^2} = \frac{12}{13}$.
Area $= \frac{1}{2} (14)(13) (\frac{12}{13}) = \frac{1}{2} (14)(12) = 84$.
Answer: $84$ cm$^2$ [3]
(1 mark for formula, 1 mark for substitution, 1 mark for answer)

15.
(a) $|\mathbf{a}| = \sqrt{3^2 + (-1)^2} = \sqrt{9+1} = \sqrt{10}$.
Answer: $\sqrt{10}$ or $3.16$ [2]

(b) $2\mathbf{a} - \mathbf{b} = 2\begin{pmatrix} 3 \\ -1 \end{pmatrix} - \begin{pmatrix} -2 \\ 4 \end{pmatrix}$
$= \begin{pmatrix} 6 \\ -2 \end{pmatrix} - \begin{pmatrix} -2 \\ 4 \end{pmatrix} = \begin{pmatrix} 6 - (-2) \\ -2 - 4 \end{pmatrix} = \begin{pmatrix} 8 \\ -6 \end{pmatrix}$
Answer: $\begin{pmatrix} 8 \\ -6 \end{pmatrix}$ [2]

(c) Vectors are parallel if $\mathbf{a} = k\mathbf{b}$.
$\frac{3}{-2} = -1.5$
$\frac{-1}{4} = -0.25$
Since $-1.5 \neq -0.25$, the ratios of corresponding components are not equal.
Thus, $\mathbf{a}$ and $\mathbf{b}$ are not parallel.
[2]
(1 mark for comparing components/ratios, 1 mark for conclusion)