Non-right-triangle surveying/navigation problems | Year 11

01 / LEARN

Understand the idea.

Translate the measurements into a model

Turn a surveying description into a labelled triangle. Mark each station, the baseline and the target; identify the required distance and its opposite angle. A convincing sketch helps organise the data but does not prove a right angle.

Bearings belong to the station where they are measured. For two directions from one station, their smaller angular separation gives an included angle. At another station, reverse the baseline bearing before comparing directions if required.

Select and explain a sequence of methods

Choose a sequence from the information available. Angle sums can complete a known side-angle pair for the sine rule; two sides and an included angle call for cosine rule. A later step may then find area or a perpendicular distance.

Keep full calculator precision between steps and state what the final distance represents: direct, horizontal or sloping. Use only assumptions justified by the description, such as a common plane, rather than inventing an instrument height or right angle.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Survey stations A and B are 30 m apart. A tree T lies on one side of AB with angle TAB=55° and TBA=70°. Find AT.

Worked solution

1
Find the target angle.
$∠ A T B = 180^{\circ} - 55^{\circ} - 70^{\circ} = 55^{\circ}$
The three interior angles of triangle ABT sum to one hundred eighty degrees.
2
Pair each side with its opposite angle.
$\frac{A T}{sin 70^{\circ}} = \frac{30}{sin 55^{\circ}}$
AT faces angle B, while the thirty-metre baseline faces angle T.
3
Calculate the required distance.
$A T = \frac{30 sin 70^{\circ}}{sin 55^{\circ}} \approx 34.4146 m$
The sine rule uses the complete baseline-angle pair.
4
Round and check the side ordering.
$A T \approx 34.4 m$
AT faces the larger angle and is therefore longer than the thirty-metre baseline.

Answer

34.4 m

EXAMPLE 02

Worked example 2

Paths from a gate follow bearings 040° and 115°. P is 18 m along the first and Q is 25 m along the second. Find PQ.

Worked solution

1
Find the included angle at the gate.
$θ = 115^{\circ} - 40^{\circ} = 75^{\circ}$
Both bearings start from the same north line at the same departure point.
2
Apply cosine rule to the direct separation.
$P Q^{2} = 18^{2} + 25^{2} - 2 (18) (25) cos 75^{\circ}$
The route lengths are two sides with their included angle known.
3
Evaluate the square and take its root.
$P Q = \sqrt{949 - 900 cos 75^{\circ}} \approx 26.7594 m$
The direct distance is a positive triangle side rather than the sum of the two route lengths.
4
Report the rounded distance.
$P Q \approx 26.8 m$
Rounding after the square root preserves one-decimal accuracy.

Answer

26.8 m

03 / YOUR TURN

Try it yourself.

Check your understanding

Two routes from the same point have bearings 025° and 100°. Find their smaller included angle.

Show answer & reasoning

75°

The smaller included angle is $100^{\circ} - 25^{\circ} = 75^{\circ}$ . The two bearings are measured from the same point and north line, so this difference is the angle between the routes.

Keep practising

6 worksheet questions · 17 marks

Open worksheet Worked solutions

Read the questions online

A 40 m baseline AB has a point P to one side. Angle $P A B = 35^{\circ}$ and $P B A = 65^{\circ}$ . Find AP.
3 marks
Routes from a depot have bearings $020^{\circ}$ and $110^{\circ}$ . Points 12 km and 16 km along them are joined. Find the joining distance.
2 marks
A triangular plot has sides 30 m and 42 m enclosing $55^{\circ}$ . Find its area to one decimal place.
2 marks
Two ground-level observers and the tower base lie on the same straight level line. Both observers are on one side of the tower, 20 m apart. The nearer elevation is $60^{\circ}$ and the farther $30^{\circ}$ . Find the tower height.
4 marks
P is 20 km from O on bearing $350^{\circ}$ . Q is 15 km from O on bearing $030^{\circ}$ . Find the smaller angle POQ and PQ.
3 marks
A survey assumes level ground, but a measured baseline slopes up a hill. Explain why its sloping length cannot automatically replace a horizontal baseline.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A 40 m baseline AB has a point P to one side. Angle $P A B = 35^{\circ}$ and $P B A = 65^{\circ}$ . Find AP.
3 marks
Angle $P = 80^{\circ}$ .
$A P = \frac{40 sin 65^{\circ}}{sin 80^{\circ}}$ .
36.8 m
Routes from a depot have bearings $020^{\circ}$ and $110^{\circ}$ . Points 12 km and 16 km along them are joined. Find the joining distance.
2 marks
The included angle is $90^{\circ}$ .
$d = \sqrt{12^{2} + 16^{2}} = 20 km$ .
20 km
A triangular plot has sides 30 m and 42 m enclosing $55^{\circ}$ . Find its area to one decimal place.
2 marks
$K = \frac{1}{2} \times 30 \times 42 sin 55^{\circ}$ .
$K = 630 sin 55^{\circ}$ square metres.
$516.1 m^{2}$
Two ground-level observers and the tower base lie on the same straight level line. Both observers are on one side of the tower, 20 m apart. The nearer elevation is $60^{\circ}$ and the farther $30^{\circ}$ . Find the tower height.
4 marks
Let x be the nearer horizontal distance. $h = x tan 60^{\circ} = (x + 20) tan 30^{\circ}$ .
Solve to obtain $x = 10 m$ .
$h = 10 \sqrt{3} m$ .
17.3 m
P is 20 km from O on bearing $350^{\circ}$ . Q is 15 km from O on bearing $030^{\circ}$ . Find the smaller angle POQ and PQ.
3 marks
The angle crosses north: $10^{\circ} + 30^{\circ} = 40^{\circ}$ .
$P Q^{2} = 20^{2} + 15^{2} - 2 \times 20 \times 15 cos 40^{\circ}$ .
$40^{\circ}$ ; $P Q \approx 12.9 km$
A survey assumes level ground, but a measured baseline slopes up a hill. Explain why its sloping length cannot automatically replace a horizontal baseline.
3 marks
A horizontal triangle uses horizontal distances.
Using an uncorrected sloping distance changes the geometry.
Its horizontal projection is shorter; the model needs a slope correction or additional height/angle information.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 General Mathematics: practical trigonometry, including similarity, right triangles, triangle area, sine/cosine rules and contextual interpretation.

Official curriculum source ↗

In this lesson

Understand the idea.

Translate the measurements into a model

Select and explain a sequence of methods

See the method in action.

Worked example 1

Find the target angle.

Pair each side with its opposite angle.

Calculate the required distance.

Round and check the side ordering.

Worked example 2

Find the included angle at the gate.

Apply cosine rule to the direct separation.

Evaluate the square and take its root.

Report the rounded distance.

Try it yourself.

Check your understanding

Keep practising