Radial Survey (HSC SSCE Mathematics Standard): Revision Notes

Worked Example: Finding Area from a Radial Survey

Question: The diagram shows a radial survey of a triangular piece of land. What is the area of $\triangle ABC$? Give your answer in square metres correct to one decimal place.

The diagram shows:

• $OA = 5$ m
• $OB = 7$ m
• $OC = 4$ m
• $\angle AOB = 82°$
• $\angle BOC = 142°$
• $\angle AOC = 136°$

Solution:

The triangular land is divided into three smaller triangles: $\triangle ABO$, $\triangle BCO$, and $\triangle ACO$. Each triangle has two known sides (the radial lines) and the included angle at $O$.

Step 1: Find the area of $\triangle ABO$

Using the sine rule for area:

$A = \frac{1}{2} \times 5 \times 7 \times \sin 82°$

$A = 17.330 \text{ m}^2$

Step 2: Find the area of $\triangle BCO$

$A = \frac{1}{2} \times 4 \times 7 \times \sin 142°$

$A = 8.619 \text{ m}^2$

Step 3: Find the area of $\triangle ACO$

$A = \frac{1}{2} \times 4 \times 5 \times \sin 136°$

$A = 6.947 \text{ m}^2$

Step 4: Add the three areas together

$\text{Total area} = 17.330 + 8.619 + 6.947$

$\text{Total area} = 32.9 \text{ m}^2$

Worked Example: Finding Distances and Perimeter

Question: A compass radial survey for a field is shown in the diagram. The radial distances and bearings are:

• $OP = 4.8$ m at bearing $072°\text{T}$
• $OQ = 4.5$ m at bearing $145°\text{T}$
• $OR = 4.7$ m at bearing $290°\text{T}$

Find: a) The size of $\angle POQ$ b) The size of $\angle POR$ c) The length of $PQ$ (to 2 decimal places) d) The length of $PR$ (to 2 decimal places) e) The perimeter of the field (to 3 significant figures)

Solution:

Part a: Finding $\angle POQ$

To find the angle at $O$ between points $P$ and $Q$, subtract the bearing of $P$ from the bearing of $Q$:

$\angle POQ = 145° - 72°$

$\angle POQ = 73°$

Part b: Finding $\angle POR$

First, work out the angle between $R$ and north:

$360° - 290° = 70°$

This means $R$ is $70°$ anticlockwise from north (or equivalently, $70°$ to the west of north).

Point $P$ is $72°$ clockwise from north.

Therefore, the angle between the two radial lines is:

$\angle POR = 70° + 72°$

$\angle POR = 142°$

Part c: Finding length $PQ$

In $\triangle POQ$, we know:

• $OP = 4.8$ m
• $OQ = 4.5$ m
• $\angle POQ = 73°$

Using the cosine rule where $PQ$ is the unknown side:

$PQ^2 = 4.8^2 + 4.5^2 - 2 \times 4.8 \times 4.5 \times \cos 73°$

$PQ^2 = 30.659...$

$PQ = \sqrt{30.659...}$

$PQ = 5.54 \text{ m}$

Part d: Finding length $PR$

In $\triangle POR$, we know:

• $OP = 4.8$ m
• $OR = 4.7$ m
• $\angle POR = 142°$

Using the cosine rule:

$PR^2 = 4.7^2 + 4.8^2 - 2 \times 4.7 \times 4.8 \times \cos 142°$

$PR^2 = 80.685...$

$PR = \sqrt{80.685...}$

$PR = 8.98 \text{ m}$

Part e: Finding the perimeter

The field is a quadrilateral with four sides: $PQ$, $QO$, $OR$, and $RP$.

Wait, let me reconsider. Looking at the diagram, the field has sides $PQ$, $QR$, $RP$ plus the radial lines aren't all sides. Actually, the field perimeter is:

$\text{Perimeter} = PQ + QO + OR + RP$

$\text{Perimeter} = 5.54 + 4.5 + 4.7 + 8.98$

Actually, reviewing the solution: the four sides of the field are $PQ + PR +$ two of the radial distances. Let me recalculate:

$\text{Perimeter} = 5.54 + 8.98 + 4.5 + 4.7$

$\text{Perimeter} = 23.72$

$\text{Perimeter} = 23.7 \text{ m (to 3 s.f.)}$