VCE General Mathematics emphasises applying geometric formulas and trigonometry in real-world scenarios. Problems typically combine several techniques and require interpretation of results.
| Context | Techniques used |
|---|---|
| Land surveying | Area, Pythagoras, trigonometry, bearings |
| Construction | Volume, surface area, scale diagrams |
| Navigation | Bearings, trigonometry, distance |
| Engineering | Scale, similarity, dimensional analysis |
| Agriculture | Area, unit conversion (ha), volume |
A triangular allotment has sides 42 m, 55 m, and the included angle between them is 68°.
Using the area formula for a triangle with two sides and included angle:
$$A = \tfrac{1}{2}ab\sin C = \tfrac{1}{2}(42)(55)\sin(68°) \approx \tfrac{1}{2}(42)(55)(0.9272) \approx 1071 \text{ m}^2$$
Convert to hectares: $\dfrac{1071}{10{,}000} \approx 0.107 \text{ ha}$.
A water pipe runs from point A to point B. B is 180 m east and 95 m north of A. The pipe must be laid in a straight line underground.
Step 1 — Length of pipe:
$$L = \sqrt{180^2 + 95^2} = \sqrt{32400 + 9025} = \sqrt{41425} \approx 203.5 \text{ m}$$
Step 2 — Bearing from A to B (clockwise from North):
$$\theta = 90° - \tan^{-1}!\left(\frac{95}{180}\right) \approx 90° - 27.8° = 62.2°$$
Bearing: 062°.
A silo consists of a cylinder (radius 3 m, height 8 m) topped with a cone (height 2 m).
$$V_{\text{cylinder}} = \pi(3)^2(8) = 72\pi \approx 226.2 \text{ m}^3$$
$$V_{\text{cone}} = \tfrac{1}{3}\pi(3)^2(2) = 6\pi \approx 18.8 \text{ m}^3$$
$$V_{\text{total}} \approx 245 \text{ m}^3$$
VCAA FOCUS: Extended response questions often describe a practical scenario requiring two or more geometric/trigonometric steps. Partial marks are awarded for each correct step, so always show working even if the final answer is wrong.
STUDY HINT: Create a personal formula sheet listing all area, volume, and trigonometry formulas. Practise recognising which formula applies from the description of a shape or situation.
