SPM Paper 2 Section C Priority (10 Marks)

Trigonometric Rules, Ambiguous Case & 3D Solids

Solution of Triangles is one of the highest scoring choices in SPM Paper 2 Section C. Master the Sine Rule, Cosine Rule, ambiguous case sketches, Heron's formula, and finding shortest perpendicular distances in 3D spatial pyramids and masts.

1. Core Mathematical Formulas

Petua Sinus (Sine Rule)

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Use when given:
• Two angles and one side ($AAS$ or $ASA$).
• Two sides and one non-included angle ($SSA$).

Petua Kosinus (Cosine Rule)

$a^2 = b^2 + c^2 - 2bc\cos A$

Rearranged for angle:
$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$
• Use for two sides and included angle ($SAS$).
• Use for three sides ($SSS$).

Area & Heron's Formula

$\text{Area} = \frac{1}{2}ab\sin C$

Heron's Formula ($SSS$):
$s = \frac{a+b+c}{2}$ (semi-perimeter)
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$

The Ambiguous Case of Sine Rule (Kes Berambiguiti)

When given two sides $a, c$ and an acute angle $\angle A$, an ambiguous case occurs if and only if: $c\sin A < a < c$.

Triangle 1 ($\triangle AB_1C$):

Acute angle: $\angle C_1 = \sin^{-1}\left(\frac{c\sin A}{a}\right) < 90^\circ$.

Triangle 2 ($\triangle AB_2C$):

Obtuse angle: $\angle C_2 = 180^\circ - \angle C_1 > 90^\circ$.

2. SPM Examiner Pitfalls in Section C

Trap 1: Degree vs Radian Calculator Mode:

Solution of triangles in SPM is exclusively solved in DEGREE mode (°). If your calculator is in RAD mode from Form 5 Circular Measure or Calculus, every single trigonometric calculation will score zero!

Trap 2: Shortest Distance from Vertex to Line:

The shortest distance from a vertex $P$ to the opposite line $QR$ is the perpendicular height $h$! Use the area equation: $\text{Area} = \frac{1}{2} \times \text{base}(QR) \times h \implies h = \frac{2 \times \text{Area}}{QR}$. Students who attempt complex coordinate geometry waste precious time and make algebraic errors.

Trap 3: Premature Rounding Off:

Do NOT round intermediate angles or side lengths to 1 or 2 decimal places. Store values in calculator memories ($A, B, C$) or keep at least 4 significant figures. Round only the final answer to 2 decimal places or 4 significant figures as required by SPM instructions.

3. Full 10-Mark SPM Paper 2 Section C Model Worked Example

SPM Paper 2 Section C Full Blueprint Model [10 Marks]
Diagram below shows a quadrilateral plot of land $ABCD$ where $ABC$ and $ACD$ are triangles.
A B C D AB = 10 m AC = 12 m BC = 8 m CD = 15 m $\angle ADC = 42^\circ$
Given that $AB = 10\text{ m}$, $BC = 8\text{ m}$, $AC = 12\text{ m}$, $CD = 15\text{ m}$, $\angle ADC = 42^\circ$, and $\angle ACD$ is obtuse.

(a) Calculate $\angle ABC$. [3 marks]

(b) Calculate $\angle CAD$. [3 marks]

(c) Calculate the total area, in $\text{m}^2$, of the quadrilateral plot of land $ABCD$. [2 marks]

(d) A vertical surveillance mast is erected at point $C$. The angle of elevation of the top of the mast from point $B$ is $35^\circ$. Calculate the shortest distance from point $C$ to the boundary line $AB$. [2 marks]

Step-by-Step Examiner Marking Scheme:
(a) In $\triangle ABC$, all 3 sides are known ($SSS$) $\implies$ use Cosine Rule for $\angle ABC$: [K1: Use Cosine Rule formula]
$\cos \angle ABC = \frac{AB^2 + BC^2 - AC^2}{2(AB)(BC)} = \frac{10^2 + 8^2 - 12^2}{2(10)(8)} = \frac{100 + 64 - 144}{160} = \frac{20}{160} = 0.125$ [K1: Correct numerical substitution]
$\angle ABC = \cos^{-1}(0.125) = 82.82^\circ$. [N1: Correct angle to 2 d.p.]
(b) In $\triangle ACD$, use Sine Rule: $\frac{AC}{\sin \angle ADC} = \frac{CD}{\sin \angle CAD}$? Wait, opposite sides: $CD$ is opposite $\angle CAD$, $AC$ is opposite $\angle ADC$. [K1: Sine Rule formulation]
$\frac{12}{\sin 42^\circ} = \frac{15}{\sin \angle CAD} \implies \sin \angle CAD = \frac{15 \sin 42^\circ}{12} = \frac{15(0.66913)}{12} \approx 0.83641$. [K1: Compute acute reference angle]
Acute angle $= \sin^{-1}(0.83641) = 56.76^\circ$. Since $\angle ACD$ is obtuse ($> 90^\circ$), $\angle CAD$ must be acute, so $\angle CAD = 56.76^\circ$. [N1: Correct $\angle CAD$]
(c) Area of $\triangle ABC = \frac{1}{2}(10)(8)\sin 82.82^\circ = 40 \times 0.99216 = 39.69\text{ m}^2$. [K1: Area $\frac{1}{2}ab\sin C$]
In $\triangle ACD$, $\angle ACD = 180^\circ - 42^\circ - 56.76^\circ = 81.24^\circ$. Area $= \frac{1}{2}(12)(15)\sin 81.24^\circ = 90 \times 0.98835 = 88.95\text{ m}^2$. [K1: Area of $\triangle ACD$]
Total Area $= 39.69 + 88.95 = 128.64\text{ m}^2$. [N1: Correct total area]
(d) Shortest distance from $C$ to line $AB$ is the perpendicular height $h_{AB}$ of $\triangle ABC$: [K1: Connect area to perpendicular height]
$\text{Area} = \frac{1}{2} \times AB \times h \implies 39.69 = \frac{1}{2}(10)h \implies 5h = 39.69 \implies h = \frac{39.69}{5} = 7.94\text{ m}$. [N1: Correct distance to 2 d.p.]
Open Form 4 Chapter 9 Section C Worksheet