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Multi-step problems and radial surveys
Problems that combine bearings with the sine and cosine rules, and compass radial surveys: finding the boundary lengths, perimeter and area of a block of land from bearings and distances measured from one central point.
This is where the chapter comes together. HSC questions on this topic are usually long: a trip on bearings, or a surveyed block of land, worth four to six marks across several parts. Each part is one of the skills from 4A to 4E. The work is in seeing which triangle to use and in what order.
A plan for every multi-step problem
- Draw it. North lines at every point where a bearing is measured.
- Find the angles inside the triangle from the bearings, using back bearings and the fact that angles around a point add to $360^\circ$.
- Ask the matching pair question from 4E: sine rule if you have a pair, cosine rule if you don't.
- Keep unrounded values in your calculator memory for the next step.
- Convert back to a bearing at the end if the question asks for one.
Bearings with the cosine and sine rules
Two fishing boats leave Port Kembla together. Boat $A$ travels $30$ km on a bearing of $040^\circ$T. Boat $B$ travels $22$ km on $155^\circ$T. Find the distance between them, and the bearing of $B$ from $A$.
A yacht sails $12$ km on $060^\circ$T, then $8$ km on $165^\circ$T. Find its distance and bearing from the start.
Compass radial surveys
To survey an irregular block, a surveyor stands at one point $O$ inside it and measures the bearing and distance to each corner. Those "radial" lines cut the block into triangles that all meet at $O$. For each triangle you know two sides (the two distances) and the angle between them (the difference of the two bearings). That is exactly SAS, so:
- each boundary comes from the cosine rule, and
- each area comes from $A = \tfrac{1}{2}ab\sin C$.
Add the boundaries for the perimeter and the areas for the total area. The angles at $O$ must add to $360^\circ$, which is a good check that you have read the bearings correctly. Land is often measured in hectares: $1$ ha $= 10\,000$ m².
Tap each row of the table to light up its triangle. Check that the angles at O add to $360^\circ$. Then move one corner's distance slider and watch which two boundaries and which two areas change, and which stay the same.
Switch to School oval. Without touching the sliders, predict which triangle has the biggest area from the angles and distances alone, then tap to check.
A surveyor at point $O$ in a paddock near Camden records: $A$ $030^\circ$T $64$ m, $B$ $110^\circ$T $80$ m, $C$ $215^\circ$T $72$ m, $D$ $300^\circ$T $58$ m. Find the area in hectares and the perimeter.
Your turn
A triangular garden is surveyed from a central point $O$: $P$ $000^\circ$T $50$ m, $Q$ $120^\circ$T $50$ m, $R$ $240^\circ$T $50$ m. Find its area to the nearest square metre.
Show the working
Each angle at $O$ is $120^\circ$ (they add to $360^\circ$). Each triangle has area $\tfrac{1}{2} \times 50 \times 50 \times \sin 120^\circ \approx 1082.53$ m². Total $\approx 3 \times 1082.53 \approx 3248$ m².
From point $O$, corner $A$ is on a bearing of $040^\circ$T and corner $B$ is on $155^\circ$T. Find angle $AOB$ in degrees.
Both bearings are measured from the same north line at $O$: $155^\circ - 40^\circ = 115^\circ$.
In a radial survey, $A$ is $050^\circ$T $40$ m and $B$ is $140^\circ$T $55$ m from $O$. Find the area of triangle $AOB$ in m².
Angle $AOB = 140^\circ - 50^\circ = 90^\circ$. Area $= \tfrac{1}{2} \times 40 \times 55 \times \sin 90^\circ = 1100$ m².
In a radial survey, $A$ is $020^\circ$T $70$ m and $B$ is $095^\circ$T $90$ m from $O$. Find the boundary $AB$ to one decimal place.
Angle $AOB = 75^\circ$. $AB^2 = 70^2 + 90^2 - 2 \times 70 \times 90 \times \cos 75^\circ \approx 9738.9$, so $AB \approx 98.7$ m.
In a compass radial survey, the total area of the block is found by:
The radial lines split the block into triangles that meet at $O$. Find each area with $\tfrac{1}{2}ab\sin C$ and add them.
Ship $X$ sails $25$ km on $015^\circ$T and ship $Y$ sails $18$ km on $110^\circ$T, both from the same port. How far apart are they, to two decimal places?
Angle at the port $= 110^\circ - 15^\circ = 95^\circ$. $d^2 = 25^2 + 18^2 - 2 \times 25 \times 18 \times \cos 95^\circ \approx 949 + 78.44 = 1027.44$, so $d \approx 32.05$ km.
A radial survey splits a block into four triangles with areas $2400$ m², $3150$ m², $2875$ m² and $1975$ m². Find the area of the block in hectares.
Total $= 2400 + 3150 + 2875 + 1975 = 10\,400$ m². Divide by $10\,000$: $1.04$ ha.
A boat sails on $060^\circ$T, then turns onto $165^\circ$T. The angle inside the triangle at the turning point is:
At the turn, the way back is the back bearing $060^\circ + 180^\circ = 240^\circ$T. Angle $= 240^\circ - 165^\circ = 75^\circ$.
Exam style: A yacht sails $12$ km on $060^\circ$T, then $8$ km on $165^\circ$T. Find its distance from the start, to two decimal places.
Angle at the turn $= 240^\circ - 165^\circ = 75^\circ$. $d^2 = 12^2 + 8^2 - 2 \times 12 \times 8 \times \cos 75^\circ \approx 208 - 49.69 = 158.31$, so $d \approx 12.58$ km.
Exam style: For the yacht in the previous question, find its bearing from the start, to the nearest degree.
Angle at the start, opposite the $8$ km leg: $\sin\theta = \dfrac{8\sin 75^\circ}{12.58} \approx 0.6142$, so $\theta \approx 37.9^\circ$. The finish is clockwise of the first leg: $060^\circ + 37.9^\circ \approx 098^\circ$T.