Worked Solutions
Measurement — Worked Solutions (HSC Maths Standard 2)
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Worked examples for HSC Maths Standard 2 measurement. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
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Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Non-right-angled trigonometry
Question
Two walking tracks leave a lookout. Track $A$ runs 1.8 km to a waterfall and track $B$ runs 2.5 km to a cave. The angle between the two tracks at the lookout is $68^\circ$.
(a) Find the straight-line distance from the waterfall to the cave, correct to one decimal place.
(b) Find the area of the triangular region enclosed by the two tracks and that straight line, correct to one decimal place.
Solution
(a) Two sides and the included angle → cosine rule. With the unknown side $d$ opposite the $68^\circ$:
$d^2 = 1.8^2 + 2.5^2 - 2(1.8)(2.5)\cos 68^\circ = 3.24 + 6.25 - 9\cos 68^\circ$.
$\cos 68^\circ \approx 0.3746$, so $d^2 \approx 9.49 - 3.371 = 6.119$, giving $d \approx 2.5$ km.
(b) Two sides and the included angle → area $= \tfrac{1}{2}ab\sin C = \tfrac{1}{2}(1.8)(2.5)\sin 68^\circ \approx 2.25 \times 0.9272 \approx 2.1$ km².
Pick the rule by what you're given: two sides plus the included angle is the giveaway for the cosine rule and the $\tfrac12 ab\sin C$ area.
(a) We know two sides and the angle squeezed between them, so the cosine rule is the right tool — it's the one that handles an included angle. The side we want, $d$, sits opposite the $68^\circ$:
$d^2 = 1.8^2 + 2.5^2 - 2(1.8)(2.5)\cos 68^\circ$. Working it through: $3.24 + 6.25 = 9.49$, and $2(1.8)(2.5)\cos 68^\circ \approx 9 \times 0.3746 = 3.371$. So $d^2 \approx 6.119$ and $d \approx 2.5$ km.
(b) For area, when you have two sides and the angle between them, use $\text{Area} = \tfrac{1}{2}ab\sin C$. Here that's $\tfrac{1}{2}(1.8)(2.5)\sin 68^\circ \approx 2.25 \times 0.9272 \approx 2.1$ km². The $\sin$ formula works because $b\sin C$ is exactly the perpendicular height of the triangle.
(a) Cosine rule (two sides + included angle):
- $d^2 = 1.8^2 + 2.5^2 - 2(1.8)(2.5)\cos 68^\circ$
- $= 9.49 - 9(0.3746) \approx 6.119$
- $d \approx 2.5$ km
(b) Area $= \tfrac{1}{2}ab\sin C$:
- $= \tfrac{1}{2}(1.8)(2.5)\sin 68^\circ$
- $\approx 2.25 \times 0.9272 \approx 2.1$ km²
Where the marks go
- 1 mark: Correct cosine rule substitution for $d$
- 1 mark: Correct distance $d \approx 2.5$ km
- 1 mark: Correct area formula $\tfrac{1}{2}ab\sin C$ with substitution
- 1 mark: Correct area $\approx 2.1$ km²
Key idea
Two sides and the included angle: use the cosine rule for the third side and $\tfrac{1}{2}ab\sin C$ for the area.
Example 2 — Rates and ratios
Question
A scale map uses a ratio of $1 : 25\,000$. A rectangular nature reserve measures 6.4 cm by 4.0 cm on the map.
(a) Find the actual length and width of the reserve in kilometres.
(b) Hence find the actual area of the reserve in square kilometres.
Solution
(a) Scale $1 : 25\,000$ means 1 cm on the map is $25\,000$ cm in reality.
Length: $6.4 \times 25\,000 = 160\,000$ cm $= 1.6$ km. Width: $4.0 \times 25\,000 = 100\,000$ cm $= 1.0$ km. (Divide cm by $100\,000$ to get km.)
(b) Actual area $= 1.6 \times 1.0 = 1.6$ km².
Convert lengths before you multiply for area — don't try to scale the area by $25\,000$, that's the classic trap.
(a) A scale of $1 : 25\,000$ tells us every 1 cm measured on the map stands for $25\,000$ cm on the ground. So we multiply each map measurement by $25\,000$.
Length: $6.4 \times 25\,000 = 160\,000$ cm. To get kilometres we divide by $100\,000$ (since 1 km $= 100\,000$ cm), giving $1.6$ km. Width: $4.0 \times 25\,000 = 100\,000$ cm $= 1.0$ km.
(b) Now that both sides are real distances, the area is just length × width: $1.6 \times 1.0 = 1.6$ km². The reason we convert first is that area scales by the square of the ratio — working with real lengths avoids that pitfall entirely.
(a) Scale $1:25\,000$, so map cm $\times 25\,000 =$ real cm.
- Length: $6.4 \times 25\,000 = 160\,000$ cm $= 1.6$ km
- Width: $4.0 \times 25\,000 = 100\,000$ cm $= 1.0$ km
(b) Area:
- $1.6 \times 1.0 = 1.6$ km²
Where the marks go
- 1 mark: Correct actual length 1.6 km
- 1 mark: Correct actual width 1.0 km
- 1 mark: Correct actual area 1.6 km²
Key idea
A scale ratio multiplies map lengths to real lengths; convert to the same unit first, then compute area from the real lengths.