3.13 Radial surveys

A surveyor wants to measure a property that is roughly in the shape of a polygon, finding its perimeter and area in preparation for its sale:

A widely used method is to make a radial survey. The surveyor picks a point inside the polygonal area and measures the distance from that point to each of the vertices of the polygon:

The lengths of the radial lines are then measured precisely, and the angles between them as well - either measuring directly, or by finding the true bearings of each radial line and finding differences. With this information the lengths of the boundary segments can be calculated using the cosine rule. The area of each triangle formed by two radial lines and a boundary segment can then be calculated using the sine area formula. The surveyor is then able to report the total length of the boundary and also the area enclosed.

 

Using the cosine rule

This diagram shows one of the triangles from the survey. We will apply the cosine rule to these numbers to find the length $c$c:

In the given triangle, we have $c^2=52^2+62^2-2\times52\times62\times\cos44.3^{\circ}$c2=522+622−2×52×62×cos44.3∘. Thus, $c^2=1933.213$c2=1933.213. Finally, by taking the square root, $c\approx43.97$c≈43.97 m, the surveyor would report a length of approximately $44$44 m for this boundary segment.

The surveyor would then perform the same calculation for each of the six triangles illustrated in the diagram. The perimeter of the polygon is then just the sum of the six boundary segments.

 

Using the sine area formula

We can now use the sine area rule to find the area of the shaded triangle.

In this case the area is given by $\text{Area}=\frac{1}{2}\times52\times62\times\sin44.3^{\circ}$Area=12​×52×62×sin44.3∘. So, $\text{Area}\approx2252$Area≈2252 m$^2$2. Notice that we do not use the length of the boundary segment, as the angle we know in this triangle is between the radial lines

The same calculation is performed for all of the triangles in the survey. So long as all the triangles formed lie completely inside the polygonal shape, the total area is found by adding the separate triangular areas. 

 

Worked example

The following diagram comes from a surveyor's field notebook. We will find the perimeter of the land and the area enclosed.

The angles of the radial lines are each measured clockwise from north. This means that the angles between the lines are the differences between the bearings: $137-49=88^{\circ}$137−49=88∘, $240-137=103^{\circ}$240−137=103∘, $322-240=82^{\circ}$322−240=82∘. The final top angle is the difference between $322^{\circ}$322∘ and north, plus $49^{\circ}$49∘, for a total of 

$\left(360-322\right)+49=87^{\circ}$(360−322)+49=87∘.

Alternatively, we can subtract the other three angles we have already found from $360^{\circ}$360∘ to complete the circle, 

$360-88-103-82=87^{\circ}$360−88−103−82=87∘.

Here is the plot of land with these angles we calculated, which we will use in the calculations that follow:

Using the cosine rule, the boundary segments have the following lengths:

$\sqrt{59^2+73^2-2\times59\times73\times\cos88^{\circ}}\approx92.25$√592+732−2×59×73×cos88∘≈92.25 m

$\sqrt{59^2+73^2-2\times59\times73\times\cos103^{\circ}}\approx103.67$√592+732−2×59×73×cos103∘≈103.67 m

$\sqrt{59^2+52^2-2\times59\times52\times\cos82^{\circ}}\approx73.01$√592+522−2×59×52×cos82∘≈73.01 m

$\sqrt{59^2+52^2-2\times59\times52\times\cos87^{\circ}}\approx76.58$√592+522−2×59×52×cos87∘≈76.58 m

The boundary has a total length of approximately $92.25+103.67+73.01+76.58=345.51$92.25+103.67+73.01+76.58=345.51 m.

From the sine area formula, the triangular pieces have the following areas:

$\frac{1}{2}\times59\times73\times\sin88^{\circ}\approx2152$12​×59×73×sin88∘≈2152 m$^2$2

$\frac{1}{2}\times59\times73\times\sin103^{\circ}\approx2098$12​×59×73×sin103∘≈2098 m$^2$2

$\frac{1}{2}\times59\times52\times\sin82^{\circ}\approx1519$12​×59×52×sin82∘≈1519 m$^2$2

$\frac{1}{2}\times59\times52\times\sin87^{\circ}\approx1532$12​×59×52×sin87∘≈1532 m$^2$2

The plot of land has a total area of approximately $2152+2098+1519+1532=7301$2152+2098+1519+1532=7301 m$^2$2.

 

Practice questions

Question 1

Question 2

Question 3