Lesson 19 ~40 min Unit 3 · Geometry +90 XP

Geometric Applications

Maths in action. Surveyors, architects, engineers and navigators all use these tools. Solve authentic problems from the real world using trigonometry and geometry.

Today's hook: The Sydney Harbour Bridge has a span of 503 m and a height of 134 m above the water. If the arch is approximately parabolic, estimate the length of the arch. What geometric concepts would you need?

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Think First

warm-up

The Sydney Harbour Bridge has a span of 503 m and a height of 134 m above the water. If the arch is approximately parabolic, estimate the length of the arch. What geometric concepts would you need? Think about whether a circular arc approximation might be easier.

Record your answer in your workbook.

The Big Idea

+5 XP to read

Trigonometry is not just abstract maths -- it is a practical toolkit used by surveyors, architects, engineers, pilots and navigators. The same formulas you have learned solve real problems every day.

The key skill in applications is translation: converting a word problem or real situation into a geometric diagram, identifying the pattern, then applying the right formula. A good diagram is half the solution.

Real world → Diagram → Formula → Verify

Draw first

A diagram reveals the pattern.

Label everything

Knowns, unknowns, bearings.

Check units

Metres, kilometres, degrees.

What You'll Master

objectives

Know

How trigonometry applies to real-world situations
Common problem types: surveying, navigation, design

Understand

How to model real situations with geometric diagrams
The importance of accuracy and appropriate rounding

Can Do

Translate word problems into geometric diagrams
Solve multi-step application problems
Evaluate the reasonableness of answers in context

Words You Need

vocabulary

SurveyingMeasuring land and creating maps using angles and distances.

TriangulationFinding a position using angles from two known points.

BearingDirection measured clockwise from North (000° to 360°).

Line of sightThe straight line from observer to object.

Back bearingThe reverse direction: add or subtract 180° from the original bearing.

Angle of elevationThe angle above the horizontal from observer to object.

Spot the Trap

heads-up

Wrong: Drawing diagrams that are not to scale or omitting important information.

Right: Always draw a clear diagram with all known information labelled. Check that your diagram matches the description.

Wrong: Giving answers with inappropriate precision. A distance of 4,217.83 m when measuring with a tape measure.

Right: Round to a sensible number of significant figures based on the precision of the given data.

Surveying Applications

+5 XP

Surveyors use trigonometry to measure distances and angles across land. Triangulation is the core technique: measure angles from two known points to locate an unknown point.

Triangulation: Measure angles from two known points to find an unknown position.
Traverse: A series of connected lines with measured lengths and angles.
Height measurement: Use angles of elevation and known distances to find heights of buildings, trees or cliffs.

The key insight: once you know two angles and one side of a triangle, the sine rule gives you everything else.

$\frac{AT}{\sin 55°} = \frac{200}{\sin 90°}$

Two angles + side

Sine rule finds all unknowns.

Find third angle first

180° minus the two known angles.

Height = distance x tan

For angles of elevation problems.

Navigation Applications

+5 XP

Pilots, sailors and hikers use bearings and trigonometry to navigate. Bearings are measured clockwise from North, from 000° to 360°.

True bearing: Clockwise from North (000° to 360°).
Back bearing: Add or subtract 180° to reverse direction.
Course correction: Calculate new bearings after drift or current.

When a journey has multiple legs, treat each leg as a side of a triangle. The angle between legs is the difference in bearings. Use cosine rule for the direct distance and sine rule for the return bearing.

$d^2 = 30^2 + 40^2 - 2(30)(40)\cos 110°$

Bearing difference

Gives the included angle.

Back bearing

Add 180° to reverse direction.

Components method

Eastings and northings as a check.

Common Pitfalls

heads-up

Skipping the diagram

Trying to solve from the words alone. Real-world problems have extra information and context that can confuse if not visualised.

Fix: always sketch a diagram. Label North, all given distances, angles and bearings.

Too many significant figures

Reporting 4,217.83 m when the given data was only precise to the nearest 10 metres. This gives a false sense of accuracy.

Fix: round to match the least precise measurement. If distances are given to 2 sig figs, answer to 2 sig figs.

Confusing bearings with angles

Using a bearing directly as an interior angle of a triangle. Bearings are measured from North, not from the previous path.

Fix: the angle between two path legs is the difference in bearings. Draw North lines at each point to clarify.

Watch Me Solve It · 3 examples

Watch Me Solve It · Triangulation

+15 XP per step

PROBLEM

Points A and B are 200 m apart. From A, the angle to a tree T is 35°. From B, the angle to T is 55°. Find the distance from A to T.

1

Draw and label the triangle

Triangle ABT with $AB = 200$ m, $\angle BAT = 35°$, $\angle ABT = 55°$

Sketch shows T is between the bearings from A and B.
2

Find angle ATB

$\angle ATB = 180° - 35° - 55° = 90°$

Conveniently a right angle.
3

Apply sine rule

$\frac{AT}{\sin 55°} = \frac{200}{\sin 90°}$

$AT = \frac{200 \times \sin 55°}{1} = 200 \times 0.819$
4

Calculate and verify

$AT \approx 163.8$ m

Check: $BT = 200 \sin 35° \approx 114.7$ m. Pythagoras: $\sqrt{163.8^2 + 114.7^2} \approx 200$ ✓

Answer$AT \approx 164$ m (2 s.f.)

Watch Me Solve It · Navigation

+15 XP per step

PROBLEM

A ship sails from port on bearing 080° for 30 km, then on bearing 150° for 40 km. Find the distance and bearing back to port.

1

Find the angle between legs

$150° - 80° = 70°$ (external angle at turn)

Internal angle = $180° - 70° = 110°$

The ship turns through 70°, so the triangle interior angle is 110°.
2

Find direct distance (cosine rule)

$d^2 = 30^2 + 40^2 - 2(30)(40)\cos 110°$

$d^2 = 900 + 1600 - 2400 \times (-0.342) = 2500 + 820.8 = 3320.8$

$d = \sqrt{3320.8} \approx 57.6$ km
3

Find return bearing (sine rule)

$\frac{\sin \theta}{30} = \frac{\sin 110°}{57.6}$

$\sin \theta = \frac{30 \times 0.940}{57.6} = 0.489$

$\theta = 29.3°$
4

Calculate bearing back to port

Back bearing of second leg = $150° + 180° = 330°$

Bearing to port = $330° - 29.3° = 300.7° \approx 301°$

Verify with components: total East = $30\sin 80° + 40\sin 150° = 49.5$; total North = $30\cos 80° + 40\cos 150° = -29.4$. Bearing = $360° - \tan^{-1}(49.5/29.4) \approx 301°$ ✓

AnswerDistance $\approx 58$ km, return bearing $\approx 301°$

Watch Me Solve It · Angles of elevation

+15 XP per step

PROBLEM

From two points 100 m apart on a straight road, the angles of elevation to the top of a tower are 25° and 40°. Find the height of the tower.

1

Set up variables

Let height = $h$, distance from first point = $x$

$\tan 25° = h/x$, $\tan 40° = h/(x-100)$
2

Equate expressions for $h$

$h = x \tan 25° = (x-100) \tan 40°$

$0.466x = 0.839(x-100)$

$0.466x = 0.839x - 83.9$
3

Solve for $x$ and $h$

$0.373x = 83.9$

$x = 225$ m

$h = 225 \times 0.466 \approx 105$ m
4

Verify

From second point: $h = (225-100) \times \tan 40° = 125 \times 0.839 \approx 105$ m ✓

Consistent. The tower is about 105 m tall.

AnswerHeight $\approx 105$ m

Copy Into Your Books

Surveying

Triangulation from known points
Two angles + side → sine rule
Height = distance x tan(elevation)

Navigation

Bearings clockwise from North
Back bearing = +/- 180°
Cosine rule for direct distance

Diagram

Always draw first
Label North, distances, angles
Mark knowns and unknowns

Check

Does the answer make sense?
Round to sensible sig figs
Use alternative method to verify

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Applications drill

4 problems

Four quick real-world problems. Draw a diagram first, then solve.

1 Two observers 50 m apart measure angles of 40° and 60° to a point. Find the distance from the first observer.

Third angle = 80°. $\frac{d}{\sin 60°} = \frac{50}{\sin 80°}$. $d = \frac{50 \times 0.866}{0.985} \approx 44$ m.$44$ m
2 A ship sails 20 km East then 15 km South. Find the bearing back to the start.

From start, ship is at bearing $126.9°$ (SE). Back bearing from ship = $126.9° + 180° = 306.9° \approx 307°$.Bearing back $\approx 307°$
3 The angle of elevation from 100 m away is 30°. Find the height.

$h = 100 \times \tan 30° = 100 \times 0.577 = 57.7$ m $\approx 58$ m.$58$ m
4 A plane flies 100 km on bearing 060°, then 80 km on bearing 150°. Find the angle between the two path legs.

External angle = $150° - 60° = 90°$. Internal angle = $180° - 90° = 90°$. The paths are perpendicular.$90°$

Complete in your workbook.

Multiple Choice · 5 questions

Triangulation

+10 XP

Two observers 50 m apart measure angles of 40° and 60° to a point. The distance from the first observer is:

25 m

35 m

44 m

55 m

C is correct. Third angle = $180° - 40° - 60° = 80°$. By sine rule: $\frac{d}{\sin 60°} = \frac{50}{\sin 80°}$, so $d = \frac{50 \times 0.866}{0.985} \approx 44$ m.

Navigation bearings

+10 XP

A ship sails 20 km East then 15 km South. The bearing from the ship back to the start is:

030°

127°

233°

307°

D is correct. From start, ship is on bearing $126.9°$ (SE). Back bearing = $126.9° + 180° = 306.9° \approx 307°$.

Angle of elevation

+10 XP

The angle of elevation from 100 m away is 30°. The height is:

50 m

58 m

67 m

100 m

B is correct. $h = 100 \times \tan 30° = 100 \times 0.577 = 57.7$ m $\approx 58$ m.

Angle between paths

+10 XP

A plane flies 100 km on bearing 060°, then 80 km on bearing 150°. The angle between the two path legs is:

60°

80°

90°

150°

C is correct. External angle = $150° - 60° = 90°$. Internal angle = $180° - 90° = 90°$. The paths are perpendicular.

Triangulation requirements

+10 XP

In surveying, triangulation requires:

One known point and one angle

Two known points and two angles

Three known points

One known distance and one angle

B is correct. Triangulation needs two known points (giving a baseline distance) and the angles from both points to the unknown position. This is the AAS pattern.

Short Answer · 3 questions

Building height

+15 XP

A surveyor measures the angle of elevation to the top of a building as 35° from point A. From point B, 50 m closer, the angle is 50°.

(a) Find the height of the building.

(b) Find the distance from B to the base.

(c) A third point C is due East of the building. From C, the angle of elevation is 25°. Find the distance from C to the base.

(a) Let height = $h$, distance from A = $x$. $h = x \tan 35° = (x-50) \tan 50°$. $0.700x = 1.192(x-50)$. $0.492x = 59.6$. $x = 121$ m. $h = 121 \times 0.700 \approx 84.7$ m.

(b) Distance from B = $121 - 50 = 71$ m.

(c) $\tan 25° = 84.7/d$. $d = 84.7/0.466 \approx 182$ m.

Ship navigation

+15 XP

A ship sails from port on bearing 080° for 40 km, then on bearing 160° for 50 km.

(a) Find the distance from port.

(b) Find the bearing from port to ship.

(a) Angle between legs = $160° - 80° = 80°$. Internal angle = $180° - 80° = 100°$.

$d^2 = 40^2 + 50^2 - 2(40)(50)\cos 100° = 1600 + 2500 - 4000 \times (-0.174) = 4100 + 696 = 4796$

$d = \sqrt{4796} \approx 69.3$ km

(b) $\frac{\sin \theta}{50} = \frac{\sin 100°}{69.3}$. $\sin \theta = \frac{50 \times 0.985}{69.3} = 0.711$. $\theta = 45.3°$.

Bearing from port = $80° + 45.3° = 125.3°$

(c) Return bearing = $125.3° + 180° = 305.3°$

Triangular park

+15 XP

A triangular park has vertices at A, B, C. $AB = 120$ m, $AC = 150$ m, angle $BAC = 55°$.

(a) Find the length of $BC$.

(b) Find the area of the park.

(d) A gardener claims the park can be divided into two equal areas by a straight path from A to the midpoint of BC. Is this true? Explain.

(a) $BC^2 = 120^2 + 150^2 - 2(120)(150)\cos 55° = 14400 + 22500 - 20664 = 16236$. $BC = \sqrt{16236} \approx 127.4$ m.

(b) Area = $\frac{1}{2}(120)(150)\sin 55° = 9000 \times 0.819 \approx 7371$ m².

(c) Path length = altitude from A = $\frac{2 \times \text{Area}}{BC} = \frac{2 \times 7371}{127.4} \approx 115.7$ m.

(d) Yes. A median from a vertex to the midpoint of the opposite side divides a triangle into two triangles of equal area, since they have equal bases (half of BC) and the same height from A.

Stretch Challenge

+25 XP

Look back at your Think First response about the Sydney Harbour Bridge. Using a circular arc model: if the span is 503 m and the height is 134 m, the radius is $R = \frac{(503/2)^2 + 134^2}{2 \times 134} = \frac{63300 + 17956}{268} \approx 302$ m. The angle subtended by half the arch is $\theta = \sin^{-1}(251.5/302) \approx 56.4°$. Find:

(a) The total angle subtended by the arch at the centre.

(b) The length of the curved arch.

(a) Total angle = $2 \times 56.4° = 112.8°$

(b) Arc length = $R \times \theta_{\text{rad}} = 302 \times \frac{112.8° \times \pi}{180°} = 302 \times 1.969 \approx 595$ m

(c) The actual arch length is approximately 595 m. If your initial estimate was between 500 m and 700 m, you were in the right ballpark. The span (503 m) is a lower bound, and the curved path must be longer.

Quick Review

all together

Draw a diagram first

Label North, all given distances, angles and bearings.

Triangulation

Two known points + two angles → sine rule finds the unknown.

Navigation

Bearings clockwise from North. Back bearing = +/- 180°.

Angles of elevation

Height = distance x tan(elevation angle).

Check significant figures

Round to match the least precise given measurement.

Verify with components

Eastings and northings as an independent check.

Interactive Navigation Challenge

+10 XP

Solve navigation problems using bearings and trigonometry. Calculate distances, bearings and positions.

Badge Wall

badges

First Step

On Fire

Perfectionist

Speedster

All Phases

Completionist

Daily Challenge

+15 XP

A hiker walks 5 km on bearing 045°, then 7 km on bearing 135°. Find the direct distance and bearing back to the start. Verify your answer using the coordinate (components) method.

Lesson Complete

+20 XP

You have finished Lesson 19 -- Geometric Applications. You can now translate real-world problems into geometric diagrams and solve them using trigonometry and geometry.

11 cards studied

4 drills completed

5 MCQs

3 SAQs

2 applications mastered