2D Trigonometry
Elevation, depression and bearings
In this note
- 6.2Use angles of elevation and depression in right-angled triangles Extended.
- 6.2Use trig with bearings: distances north and east, and the bearing between two points.
- 6.2Know that the shortest distance from a point to a line is the perpendicular Extended.
Before you start: Pythagoras and right-angled trigonometry and bearings.
Formula bank
Angles of elevation and depression Extended
Look up from the ground: the angle between the horizontal and your line of sight is the angle of elevation. Look down from the top: the angle below the horizontal is the angle of depression. The two are equal (alternate angles), so you can always work in the triangle on the ground.
General methodHow to answer any elevation or depression question
- Draw the right-angled triangle: ground, vertical height, line of sight.
- Put the angle at the bottom, between the ground and the line of sight (a depression angle moves there too).
- Height is opposite the angle, ground distance is adjacent: use tan.
- Two angles from two places? Find each ground distance, then subtract.
Example From a point 40 m from the foot of a tower, the angle of elevation of the top is 32°. Find the height.
opp and adj: tan.
- \(h=40\tan32^\circ=24.99\ldots\)
Answer 25.0 m
Example From the top of a 60 m cliff, the angle of depression of a boat is 25°. How far is the boat from the foot of the cliff?
Move the 25° down to the boat. Height 60 is opposite, distance is adjacent.
- \(\tan25^\circ=\dfrac{60}{d}\)
- \(d=\dfrac{60}{\tan25^\circ}=128.7\)
Answer 129 m (3 s.f.)
Example A building is 30 m tall. From point P the elevation of the top is 20°; from Q, nearer, it is 35°. Find PQ.
Two ground distances, then subtract.
- \(\dfrac{30}{\tan20^\circ}=82.42\)
- \(\dfrac{30}{\tan35^\circ}=42.84\)
- \(PQ=39.58\)
Answer 39.6 m
Try it yourself
Show answer
Answer 20.6
Show answer
Answer 129
Bearings with trigonometry
A bearing is an angle measured clockwise from north. Drop a right-angled triangle from the journey: one side points north, one points east. Then use SOH CAH TOA as normal.
General methodHow to answer any bearing calculation
- Draw a north line at the starting point.
- Make a right-angled triangle with the north and east directions.
- Use Pythagoras for the distance and tan for the angle from north.
- Turn the angle into a bearing: clockwise from north, three figures. Need the bearing back? ± 180°.
Example B is 8 km east and 6 km north of A. Find the distance AB and the bearing of B from A.
The angle θ at A is measured from north, so the east side is opposite.
- \(AB=\sqrt{8^{2}+6^{2}}=10\)
- \(\theta=\tan^{-1}\dfrac86=53.1^\circ\)
Answer 10 km on a bearing of 053.1°
Example A ship sails 12 km on a bearing of 060°. How far east and how far north is it from its start?
Angle from north 60°: east is opposite, north is adjacent.
- east \(=12\sin60^\circ=10.39\)
- north \(=12\cos60^\circ=6\)
Answer 10.39 km east, 6 km north
Try it yourself
Hint
Show answer
Answer 202.6
The shortest distance to a line Extended
The shortest way from a point to a straight line is the line that meets it at right angles. So 'the shortest distance from C to the path AB' means: drop a perpendicular CX, then use the right-angled triangle ACX.
Example AC = 20 m and angle CAB = 35°. Find the shortest distance from C to the line AB.
CX is perpendicular, so triangle ACX has a right angle at X. CX is opposite 35°.
- \(CX=20\sin35^\circ=11.47\)
Answer 11.5 m
Ways they test it
Type 1 Height from elevation
"The angle of elevation of the top of the tree is 32°. Find its height." (2 marks)
Spot it 'elevation', a distance along the ground.
- Height = distance × tan(angle).
Type 2 Depression
"From the top of the lighthouse the angle of depression of the boat is 25°." (3 marks)
Spot it 'depression', a height.
- Move the angle to the bottom (alternate).
- distance = height ÷ tan(angle).
Type 3 Two observers
"Find the distance between the two points." (4 marks)
Spot it two angles of elevation of the same top.
- Each ground distance separately.
- Subtract (same side) or add (opposite sides).
Type 4 Bearing and distance
"Calculate the bearing of B from A." (3 marks)
Spot it distances north/east, or a triangle with a north line.
- tan⁻¹ for the angle from north.
- Convert to a three-figure bearing.
For example 8 km east, 6 km north
- \(\tan^{-1}(8/6)\)
Answer 053.1°
Type 5 Shortest distance
"Calculate the shortest distance from C to AB." (3 marks)
Spot it 'shortest distance' to a line.
- Perpendicular from C.
- Right-angled trig.
Mistakes to avoid
- ✗ Depression angle at the top used inside the triangle✓ move it to the bottom (alternate angles)The angle at the top is between the HORIZONTAL and the line of sight.
- ✗ Bearing given as 53.1°✓ 053.1°Bearings have three figures before the decimal point.
- ✗ Measuring the bearing from east✓ always clockwise from north
- ✗ Shortest distance along a slanting line✓ it is the perpendicular
Summary
- Elevation looks up, depression looks down; both from the horizontal, and they are equal.
- Height = distance × tan(angle).
- Bearings: north line, right-angled triangle, tan⁻¹ for the angle, three figures.
- Shortest distance to a line = perpendicular.
If you see …
- 'elevation' or 'depression'right-angled triangle on the ground, tan.
- 'bearing of'north line at the FROM point.
- 'shortest distance'perpendicular.
Practice
Practice: easy, medium, hard
Easy
Show answer
Answer 233
Show answer
Answer 25.0
Show answer
Answer 10
Medium
Guide me
What is it asking? a distance
What do we know? 60 opposite, d adjacent
Plan \(60\div\tan25^\circ\)
Show working
- 128.7
Answer 129
Guide me
What is it asking? a distance
What do we know? east is opposite 60°
Plan \(12\sin60^\circ\)
Show working
- 10.39
Answer 10.39
Guide me
What is it asking? a bearing
What do we know? east opposite, north adjacent
Plan \(\tan^{-1}(8/6)\)
Show working
- 053.1°
Answer 53.1
Hard
Guide me
What is it asking? a distance
What do we know? two ground distances
Plan subtract
Show working
- 82.42 − 42.84
Answer 39.6
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(30\div\tan20^\circ\) | M1 | ||
| \(30\div\tan35^\circ\) | M1 | ||
| 39.6 | A1 | ||
| Total: 3 marks | |||
Guide me
What is it asking? a bearing
What do we know? Q from P: \(\tan^{-1}(5/12)=22.6^\circ\)
Plan back bearing: + 180
Show working
- 202.6°
Answer 202.6
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(\tan^{-1}(5/12)\) | M1 | ||
| +180 | M1 | ||
| 202.6 | A1 | ||
| Total: 3 marks | |||
Guide me
What is it asking? a distance
What do we know? perpendicular CX
Plan \(20\sin35^\circ\)
Show working
- 11.47
Answer 11.47
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(20\sin35^\circ\) | M1 | ||
| 11.47 | A1 | ||
| Total: 2 marks | |||
Can I do this?
- I can use angles of elevation and depression.
- I can find distances and bearings with trigonometry.
- I know the shortest distance to a line is the perpendicular.
Anything unticked: revise that section first. All ticked: take the quiz.
Quiz
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